Skip to main content

Full text of "Engineering applications of higher mathematics"

See other formats

This is a digital copy of a book that was preserved for generations on library shelves before it was carefully scanned by Google as part of a project 
to make the world's books discoverable online. 

It has survived long enough for the copyright to expire and the book to enter the public domain. A public domain book is one that was never subject 
to copyright or whose legal copyright term has expired. Whether a book is in the public domain may vary country to country. Public domain books 
are our gateways to the past, representing a wealth of history, culture and knowledge that's often difficult to discover. 

Marks, notations and other marginalia present in the original volume will appear in this file - a reminder of this book's long journey from the 
publisher to a library and finally to you. 

Usage guidelines 

Google is proud to partner with libraries to digitize public domain materials and make them widely accessible. Public domain books belong to the 
public and we are merely their custodians. Nevertheless, this work is expensive, so in order to keep providing this resource, we have taken steps to 
prevent abuse by commercial parties, including placing technical restrictions on automated querying. 

We also ask that you: 

+ Make non-commercial use of the files We designed Google Book Search for use by individuals, and we request that you use these files for 
personal, non-commercial purposes. 

+ Refrain from automated querying Do not send automated queries of any sort to Google's system: If you are conducting research on machine 
translation, optical character recognition or other areas where access to a large amount of text is helpful, please contact us. We encourage the 
use of public domain materials for these purposes and may be able to help. 

+ Maintain attribution The Google "watermark" you see on each file is essential for informing people about this project and helping them find 
additional materials through Google Book Search. Please do not remove it. 

+ Keep it legal Whatever your use, remember that you are responsible for ensuring that what you are doing is legal. Do not assume that just 
because we believe a book is in the public domain for users in the United States, that the work is also in the public domain for users in other 
countries. Whether a book is still in copyright varies from country to country, and we can't offer guidance on whether any specific use of 
any specific book is allowed. Please do not assume that a book's appearance in Google Book Search means it can be used in any manner 
anywhere in the world. Copyright infringement liability can be quite severe. 

About Google Book Search 

Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers 
discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web 

at |http : //books . google . com/ 




Digitizea l 





Digitized by LjOOQ IC 

Digitized by LjOOQ IC 



The Electric Circuit 

XV + 229 pages, 6 by 9. Cloth $2.00 

(Second Edition, entirely rewritten and er^arged) 

The Magnetic Circuit 

xviii + 283 pages, 6 by 9. Cloth $2.00 

Published by JOHN WILEY & SONS, Inc. 

Experimental Electrical Engineering 

Vol. I. xix + 469 pages, 6 by 9, 328 figures. 

Cloth. $3.50 net 

Vol. II. xiv + 333 pages, 6 by 9, 209 figures. 

Cloth $2.50 net 

Engineering Applications of Higher Math- 

Part I. Machine Design, xiv + 69 pages, 

6iby8. Cloth $0.75 net 

Part II. Hydraulics, v + 103 pages, SJ 

by 8. Cloth..... $0.75 net 

Part III. Thermodynamics. v + 113 pages, 

6iby8. Cloth $0.75net 

Part rV. Mechanics op Materials. 

V + 81 pages, 6i by 8. Cloth $0.75 net 

Part V. Electrical Engineering, vii + 

65 pages, SJ by 8. Cloth $0.75 net 

Elementary Electric Testing 
Loose Leaf Laboratory Manual of the Wiley 
Technical Series, J. M. Jameson, Editor. 
25 direction sheets with numerous diagrams 
and cuts. Complete in removable paper 
cover $0.50 net 


Ueber Mehrphasige Stromsysteme bei 
Ungleichmassiger Belastung. Paper Mk. 2.40 

Digitized by LjOOQ IC 












Lohooh: CHAPIIAN & HALL, Luuibd 

Digitized by LjOOQ IC 





Copyright. 1916, 



Stanbope pttn 


Digitized by LjOOQ IC 


This last part of the work contains problems in elec- 
trical engineering, and these, as well as the problems 
of the other parts, have been selected to illustrate the 
important principles of analytics and calculus. The 
several parts are independent of one another, and con- 
tain problems selected from the following engineering 

Part I. Machine Design. 

Part II. Hydraulics. 

Part III. Thermodynamics. 

Part IV. Mechanics of Materials. 

Part V. Electrical Engineering. 

A student or an engineer who wishes to review cal- 
culus or analytic geometry, or who wishes to acquire 
facility in appUcations of higher mathematics to en- 
gineering problems, may begin with any one of the 
five parts. He will probably first read the part which 
deals with his own branch of engineering, and then 
take the others with which he is less famihar. In this 
way every reader can arrange the parts according to 
an ascending degree of difficulty. 

The author's views on teaching mathematics to en- 
gineering students may be found in the Preface to 
Part I and in the Dialogue following that preface. 
Reference may also be made to Part I for a list of 
works on mathematics and for an Appendix entitled, 
''What a Senior in Engineering ought to know about 

With this fifth part is completed a work which rep- 
resents the result of many years of study and effort 
on the part of the author to inspire his fellow engineers 


Digitized by VjOOQ IC 


with love for mathematics and to unpress upon them 
the simplicity, the importance, and the efficacy of the 
mathematical method. In 1898, John Perry's ''Cal- 
culus for Engineers," came to the author's notice and 
since then he has agreed with the writer that in teach- 
ing analytics and calculus to engineering students the 
instructor should Umit himself at first to a compara- 
tively small number of curves, functions, and trans- 
formations, but that each of these must be explained 
and made concrete in a number of engineering prob- 
lems carried through to their practical conclusions. 

A natural mathematician enjoys abstract relation- 
ships from which the concrete appUcations are ob- 
tained as special cases. But most engineers are not 
natural mathematicians, and they learn mathematics 
much better by beginning with the more specific cases 
and gradually generaUzing them as the number of 
these cases increases. Such is their type of mind, a 
type that is valuable in practical work, and it is im- 
psychological to train them in the opposite way. It 
would seem that either our instructors in mathematics 
must come to understand this difference between their 
temperament and that of their students, or else mathe- 
matics in engineering colleges will have to be taught 
by engineers. 

Another important and much needed improvement 
in teaching mathematics to our engineering students 
would be to carry the subject through all the four or 
five imdergraduate years, and not discontinue it at 
the time when the student is just beginning to imder- 
stand what it is all about. I beUeve it was the emi- 
nent electrician Dr. C. P. Steinmetz, who expressed 
the correct psychological idea that a scientific method 
which is worth studying at all is worth continuing 
throughout the college course; he meant by this that 

Digitized by VjOOQ IC 


the threads of thought once grasped shoidd not be 
allowed to break. 

Combmmg therefore the two foregoing reqmrements, 
namely, from the specific to the general, and the con- 
tinuity of training, we get the following outline of a 
rational course in mathematics for engineering students : 

First Year. AppUcations of elementary mathematics 
to engineering problems; additional chapters on soUd 
geometry and trigonometry, with appUcations. 

Second Year. The straight line and a few simple 
curves in analytics; differentiation and integration of 
a few simple functions; numerous applications to en- 
gineering problems. 

Third Year. Elementary analytics and calculus, 
with appUcations to engineering and physics. 

Fourth Year. Advanced chapters from analytics and 
calculus; generaUzation of the information previously 
gained, with more stringent proofs; a few simple 
differential equations. 

Fifth Year (Optional). Differential equations, 
hyperbolic functions, theory of probabihties, or any other 
branch of higher mathematics that the student is inter- 
ested in or that may be of use to him in his specialty. 

Such an arrangement of the subject matter may be 
called concentric, because it may be symbolically rep- 
resented by a series of concentric circles, with their 
zones covering more and more ground, and each ex- 
tending in all directions.* 

Of course, a continuous schedule in mathematics 
would mean some rearrangement in the order of the 
other studies, but such a rearrangement is highly de- 
sirable anyway, if one accepts the principle of leading 

* See the author's paper " On the Concentric Method of Education 
in Engineering," Proc. Soc. Promotion Eng'g Education, Vol. 16 (1908), 
p. 258. 

Digitized by VjOOQ IC 


from the specific and the concrete, to the general 
and the abstract. The descriptive courses in engineer- 
ing, as well as shop, drafting, and simple laboratory 
experiments, would have to be more concentrated in 
the first two years of the course, and the more theo- 
retical courses taught later on the basis of the tan- 
gible engineering information already acquired. In 
particular, physics and mechanics with their general 
laws ought to be shifted towards the end of the course 
when the student is mature enough to appreciate 
them. A few simple experimental laws of nature 
which are used in engineering can be easily given in 
the corresponding technical courses. As a matter of 
fact, teachers of engineering actually find it necessary 
to repeat these laws to their students in spite of the 
preceding courses in physics and mechanics. 

Originally Part V was planned to be much larger, 
but in the meanwhile the author has published two 
other books of a mathematical character, "The Elec- 
tric Circuit" and "The Magnetic Circuit" (McGraw- 
Hill Book Co.), and rather than to duplicate the text 
he limited himself in this book to electrical problems of 
a somewhat different type. The principal difficulty in 
selecting problems from electrical engineering for a book 
of this character consists in avoiding relationships the 
physical significance of which may not be clear to the 
student. Since we have no electric or magnetic sense 
among the five external senses, these relations are boimd 
to remain more intangible than the phenomena dealt 
with in mechanics, hydraulics, thermodynamics, or 

Mr. A. C. Stevens, M.E., instructor in Cornell Uni- 
versity, has read the proofs of this volmne and to him 
the author is imder obligation for the care with which 
the printing was done. 

GoBNBLL Uniybbsitt, Ithaca, N. Y. 
Mayy 1916. 

Digitized by LjOOQ IC 


Chapter Pao« 
I. Resistance of, Non-Ctlindrical Conductobs and Mag- 
netic Paths 3 

II. MiNiBfUM Weight of Metal in a Combination of Con- 
ductors 15 

III. Most Economical Size op Conductors 22 

IV. A Leaky Direct-current Line 37 

V. Average and Effective Values of Voltages and 

Currents • 43 

VI. Efficiency and Losses in Electrical Machinery 61 

Appendix 62 

List of Reference Works 65 

Digitized by LjOOQ IC 

Digitized by LjOOQ IC 



Digitized by LjOOQ IC 

Digitized by LjOOQ IC 




The three fundamental quantities with which an 
electrical engineer has constantly to deal are: electro- 
motive force, current, and electrical resistance. The 
intrinsic nature of electrical phenomena being im- 
known, it is hardly possible to convey to a beginner a 
clear idea of these quantities, though to an electrical 
engineer or a physicist these quantities become real 
through constant usage. Assimiing the flow of elec- 
tricity to be analogous to the flow of a liquid through 
a pipe, an electric current may be compared to the 
quantity of the Uquid passing per second through a cross 
section of the pipe. The resistance of the conductor 
may be likened to the friction between the fluid and the 
pipe. The electromotive force, or the difference of 
potential between two cross sections of a conductor, 
is then analogous to the difference in pressure between 
the corresponding cross sections of the pipe; this loss 
of pressure being necessary for overcoming the friction 
along the pipe. 

Or, comparing the flow of electricity to that of heat 
along a rod, the electromotive force corresponds to the 
difference of temperatures, being the cause of the 
flow; the current is analogous to the quantity of heat 


Digitized by LjOOQ IC 


flowing through a cross section of the rod per unit 
time; the electric resistance corresponds then to the 
resistance of the rod to the flow of heat; in other words, 
it is the reciprocal of the thermal conductance of the 

Ohm foimd experimentally that for a given metaUic 
conductor, at a constant temperature, the current is 
proportional to the appUed electromotive force. For 
instance, let an electromotive force of 10 volts between 
the ends of a conductor produce a current of 4 amperes 
in the conductor. Then an electromotive force of 
20 volts will cause a current of 8 amperes to flow. 
This relation is known as Ohm's law, and is generally- 
written in the form 

^=227, ...... (1) 

where E is the difference of potential, or the voltage 
between the ends of the conductor, / is the current and 
R is the resistance of the conductor, or the coefficient 
of proportionality between E and /. 

With long cylindrical rods, strips, wires, and in gen- 
eral where electricity flows in parallel stream lines 
uniformly distributed throughout the cross section of a 
conductor, the resistance is proportional to the length 
I of the conductor and is inversely proportional to 
its cross section g. Thus, 

fi = P-^, (2) 

and Eq. (1) becomes 

JS = p.-/ (3) 

In these equations p is a coefficient which depends 
upon the material of the conductor and upon its tem- 

Digitized by VjOOQ IC 

Chap.ij non-cylindrical conductors. 5 

perature. The coefficient p is called the resistivity of 
the material. 

For copper and aluminum wires, and for some other 
metals commonly used as electrical conductors, tables 
are available in various handbooks from which resist- 
ance may be readily obtained for standard cross sec- 
tions and lengths. Thus, for instance, the resistance 
of a copper wire having a cross section of 1 sq. nmi., 
and 1 meter long, at a standard temperature of 25'' C, 
is about 0.018 ohm. This is also the value of p for 
copper at this temperature, provided that lengths are 
measured in meters and cross sections in square milli- 
meters, because then in Eq. (2) Z = 1 and ? = 1. 
Therefore, for a trolley wire 3 km. long and of 70 sq. 
mm. in cross section, the resistance is, according to 
Eq. (2), 

R = 0.018 X 3000/70 = 0.77 ohm. 

When a current of, say, 40 amperes is flowing through 
this wire the voltage across its ends is 40 X 0.77 = 30.8 

Ohm's law may be expressed in words by saying that 
in a metal conductor the electrical effect (current /) 
is proportional to the cav^e (voltage E). There are 
some other natiu-al phenomena in which the effect is 
proportional to the cause; for instance, 

(1) In a sand filter, imder certain conditions, the 
quantity of the Uquid which passes through the sand is 
proportional to the appUed hydraulic pressing or head. 

(2) In a homogeneous wall, the quantity of heat 
which passes across it in irnit time is proportional to 
the difference in temperature of the two surfaces. 

(3) In the absence of iron, the magnetic flux pro- 
duced by a current in a given coil is proportional to 
that current. 

Digitized by LjOOQ IC 


(4) In dielectrics (commonly known as insulating 
substances) a displacement of electricity, or dielectric 
flux, is proportional to the voltage which produces it. 

For all these cases equations may be written which 
are mathematically identical with Eqs. (1), (2) and 
(3), although their physical significance is entirely 
different. Some authors even speak of Ohm's law in 
application to the flow of heat, or to magnetic phenom- 
ena, etc. Therefore, the problems solved in this chap- 
ter apply to several other classes of phenomena than that 
of the flow of current through metallic conductors, and 
the student will do well to make clear to himself the 
physical significance and the practical importance of 
the results in appUcation to these different phenomena. 

In the solution of the problems that follow, the 
student will need the following theorem: When two 
or more resistances are connected in series the equiva- 
lent resistance is equal to the sum of the component 
resistances; or 

fieq = Si? (4) 

A proof of this proposition will be found in almost any 
elementary text-book on electricity.* 

In some problems resistances are connected in par- 
allel; in this case it is convenient to consider the recip- 
rocal of each resistance, called the conductance. We 
have, from Eq. (2), the following expression for the 
conductance fir of a conductor: , 

'-i---f w 

where y is equal to 1/p and is called the conductivity of 
the material. By analogy with Eqs. (1) and (3) we have : 

I = Eg, (6) 

* See, for instance, the author's "Electric Circuit," p. 7. 

Digitized by VjOOQ IC 

Chap. I.] 



' - -f «■ 


The following theorem applies to conductances in par- 
allel: When two or more conductors are connected in 
parallel, the equivalent conductance is equal to the 
sum of the component conductances;* or 

(7eq = S? (8) 

As a general rule, the student will find it conven- 
ient to use resistances when conductors are connected 
in series, and to use conductances when conductors are 
connected in parallel. 

-Q--.— ■ 

Fig. 1. — A wedge-shaped conductor. 

Prob. I. — Find the resistance of a wedge-shaped conductor 
(Fig. 1) of length I and width /, the height at one end being a, and 
at the other b. Solution, — Let the traces of the two converging 
planes meet at 0, at a distance xo from a. Consider the conductor 
to consist of infinitesimal prismatic layers, such as MN, If y is 
the height of such a layer, its resistance is p dx/(fy). According to 

* For a proof of this proposition see, for instance, the author's 
"Electric Circuit," p. 8. 

Digitized by VjOOQ IC 


Eq. (4), the total resistance of the conductor is equal to the sum of 
the resistances of the infinitesimal slices, or 

R= C'pdx/ify) (9) 

In order to be able to integrate this equation, y must be expressed 
as a fimction of x. We have from the figure that xq : a = x : yy 
so that y = ox/xq. Substituting this value in Eq. (9) and inte- 
grating we find that 

R = ^Laf, (10) 

fa xq 

where the symbol Ln stands for the natural logarithm. The 
constants Xq and Xi may be eliminated by using the conditions 
Xi/xo = 6/a, and xq/I = a/{b — a). 

Prob. 2. — Find the resistance of a conductor shaped like a 
truncated cone, the length of the axis being Z, and the radii of the 
extreme cross sections ro and n. Hint. dR =^ p dx/(rT^), where 
the variable r must be expressed through x from the similar tri- 
angles, as in Prob. 1. 

Prob. 3. -^ On the basis of the answer to Prob. 2, show that the 
resistance of a tapering circular conductor is equal to that of a 
cylindrical conductor of the same material and length, the cross 
section of the cylinder being the geometric mean of the extreme 
sections of the cone. 

Prob. 4, — A conductor has the shape of a hollow truncated 
cone, the thickness of the wall being the same at all places; De- 
duce an expression for its resistance. 

Prob. 5. — A conductor has the shape of a hollow truncated 
cone; the radii at one end are tq and n, and at the other end r2 and 
rs. Find an expression for its resistance. 

Prob. 6. — A conductor is formed by the revolution of an ordi- 
nary parabola (Fig. 2) about axis OX. The extreme cross sections 
are qo and gi, the length is I. What is the resistance of the con- 
ductor? Hint, yo^ = qo/Tr = 2 pxo; 2/1* = gi/x = 2 pxi; y^ = 
q/tr = 2tx; Xi — xo = l. 

Prob. 7. — A conductor has the shape of a body of revolution 
formed by the rotation of a given curve y = f (x) about the X-axis. 
Write down an expression for the electrical resistance of the con- 
ductor between two given cross sections. 

Digitized by LjOOQ IC 



Prob. 8. — Prove that the curve of voltage gradient Gx in Prob. 1 
is a hyperbola. The voltage gradient is defined as the voltage drop 
per unit length of the conductor. Solviion. — The resistance of 
an infinitesimal layer being p dx/{fy)j the voltage drop across this 

Fig. 2. — A conductor formed by revolution of a parabola. 

layer is de = ip dx/(Jy)y where i is the current flowing through the 
conductor. By the definition of the voltage gradient, Gx = de/dx = 
ip/{fy) . Substituting the value of y found in the solution of Prob . 1 , 
we find Gx = ipxo/{fax)y which is a hyperbola of the form Gx'X = 

Prob. 9. — Find the equations of the curves of voltage gradient 
in Prob. 2 and in Probs. 4 
to 7. 

Prob. ID. — Find the re- 
sistance of the insulation of 
a single-core concentric cable 
(Fig. 3) [of length Z. Solu- 
tion. — Let the core be con- 
nected to the positive pole 
and the sheath or lead cover 
to the negative pole. The 

insulation being imperfect, ^ ^ * . , . • , , 

,, . i. J r i.u Fig. 3. — A smgle-core concentric cable, 

there is a tendency for the ^ 

current to flow from the core into the sheath. The resistance of 

the insulation may be resolved into the resistances of infinitesimal 

concentric cylindrical layers. Consider such a layer of radius x 

and of thickness dx. According to formula (2), the resistance 

Xead i^eajli 


Copper caro 

Digitized by VjOOQ IC 


of this shell is dR = p dz/(2irxl). Integrating this expression 
between the limits^a and b, we find R = (p/2 tI) Ln (b/a)* 

Prob. II. — The insulation of the cable shown in Fig. 3 consists 
of two concentric layers of resistivities pi and P2 respectively. The 
separating surface between the two is of radius c. Write down an 
expression for the resistance of the whole insulation. 

Prob. 12. — Referring to Prob. 10, if the total voltage between 
the core and the sheath is E, what is the voltage e between two 
points in the insulation at distances n and r2 from the center? 
Hint, de = ip dx/{2 irxZ), where i is the unknown leakage cur- 
rent. Integrate this expression between the limits n and r2, and 
eliminate the unknown current using the condition that e = E, 
when ri = a and r2 = b. 

Prob. 13. — The diameter of the core of a concentric cable is 
9 mm., the thickness of the insulation is 4 mm.; the leakage current 
per km. length is i. Can the leakage current be reduced to J i by 
increasing the thickness of the insulation to 8 mm.? 

Prob. 14. — Let Fig. 3 represent concentric spheres, instead of 
cylinders. Solve Probs. 10 to 13 for this case. 

Prob. 15. — The final answer to Prob. 1 is of the form 

fi = Const. X?f^^ (11) 

For a conductor of constant thickness, 6 = a, and the foregoing 
expression assumes an indeterminate form 0/0. Show that the 
true value of the ratio is 1. Hint, — Let b = (1 + e) a, where 9 
is a small quantity, and evaluate the indeterminate form according 
to the rule given in the differential calculus. 

Prob. 16. — When b in formula (11) differs but little from a, 
numerical computations become unsatisfactory, because a small 
error in either value may affect the result to a considerable degree. 
Show that it is better in this case to expand R into an infinite 
series. Hint, — Expand Ln (1 + 0) in powers of B, and divide each 
term by 0, 

Prob. 17. — An electric current flows between electrodes A and 
B (Fig. 4) through a medium of comparatively high resistivity. 
Electrode A consists of " teeth," electrode B is a large mass of 

* A similar method may be used in deriving the electrostatic capacity 
of a concentric cable; see the author's "Electric Circuit," Chap. XVI. 

Digitized by VjOOQ IC 

Chap.ij non-cylindrical conductors. 


metal limited by a plane. Both electrodes are assumed to extend 
indefinitely in the direction perpendicular to the plane of the paper, 
so that the problem is a two-dimensional one. Determine the 
resistance of the medium between the electrodes, within the limits 




II 1 1 


c t 


< [8— 







Fig. 4. — Stream lines between teeth and a solid pole. 

MN and ilf W, for a length I in the direction perpendicular to the 
plane of the paper. The stream lines, or the lines of flow of current, 
are assmned to consist of straight lines and concentric quadrants.* 
SoliUion, — Since the individual stream lines are electrically con- 
nected in parallel, it is more convenient to consider their conduct- 
ances, according to Eqs. (5) and (8). In the region between MN 
and PQ the conductance is ^« = 7 • i tl/a, where the subscript t 
stands for " tooth." In the region between PQ and M'N' consider 
an infinitesimal conductance of width dx corresponding to the 
quadrant of radius x. Its conductance is dg, = yl dx/{\ ttx + a), 
where the subscript s stands for slot. Integrating between the 
limits X = and x = is we find g, = (2 yl/v) Ln (xs/4 a + 1). 

* This and the remaining problems in the chapter have a direct 
application in magnetic calculations of electro-dynamic machinery. 
Thus, electrode B may represent a magnetic pole-piece, and electrode A 
part of a slotted armature. The required resistance corresponds to 
the reluctance of the air-gap. 

Digitized by VjOOQ IC 




The total conductance g = gt + g», and consequently the required 
resistance R = l/(gt + g,)- 

Piob. 18. — In Prob. 17 let the teeth of the electrode A be circular 
cylinders of diameter t. Determine the resistance of the medium 
corresponding to one tooth assuming the same shape of stream lines. 

Prob. 19. — Two flat electrodes of width h (Fig. 5) extend indefi- 
nitely in the direction perpendicular to the plane of paper. They are 






/ ^^ 



^^"^\ \ 

' /y^"^^^ 






""■^^V \\ 


\ ^^ V\\ ^ 

r^ /''^ ' 

\ ^\ \ \ \ \ 

i \ ( i r 


'^ i » 1 1 1 



t h ^ 


* - > 

Fig. 5. — Two assumed systems of lines of flow of current between 
flat electrodes. 

surrounded by a medium of low conductivity, and the stream lines 
of the current are assumed to consist of straight lines connected by 
concentric quadrants (full drawn lines). Deduce an expression for 
the conductance of the medium over a length I in the direction per- 
pendicular to the plane of the paper. 

Prob. 20. — Solve Prob. 19 assuming the lines of flow to be con- 
centric semi-circles, as shown in Fig. 5 by dotted lines. 

Prob. 31. — The stream lines between electrodes A and B (Fig. 
6) have a general shape shown in the figure. The minimum dis- 
tance between the electrodes is a, the maximum distance, measured 
along a line of flow, is a + Aa. The lower electrode is so propor- 
tioned that the increase in the length y of lines of flow is proportional 
to the square of the distance x from the center. Calculate the 
conductance of the medium for a length I in the direction perpen- 
dicular to the plane of the paper, and determine the equivalent 

Digitized by VjOOQ IC 

Chap. I.] 



distance Oeq between two flat electrodes which would give the 
same conductance.* Hint, t/ = a + Aa (x/wy; the infinitesimal 
conductance \a dg = yl dx/y. The distance Oeq is determined from 
the condition g = ylw/Oeq. 

a + Aa 

Fig. 6. — A gap of variable length between two electrodes. 

Prob. 22. — Solve Prob. 21 when the increase in the length y is 
proportional to the nth power of the distance x from the center. 


! n 

1 J^ 

_1_ jfe 

' M 



1^ -T.^^ J 

Fig. 7. — Lines of flow of current between two incUned planes; three 
simple assumptions. 

Prob. 23. — Two electrodes (Fig. 7) have plane surfaces inclined 
at an angle 2 9 to one another. It is required to calculate the con- 
ductance of the medium between them, assuming in succession the 
lines of flow to be shaped as shown in (a), (6) and (c). 

* For a practical application of this problem see the author's " Mag- 
netic Circuit, " pp. 91 and 92. 

Digitized by VjOOQ IC 


Prob. 34. — In nature the actual flow of current takes place in 
such a way that the conductance of the path is a maxiTnum. Show 
that assumption (a) in Fig. 7 is more correct than (6), and that 
(6) is more correct than (c). Hint. — Compare the conductances 
of the coiresponding infinitesimal paths, remembering that tan > 
$> mie. 

Digitized by LjOOQ IC 



Many engineering problems admit of a large number 
of solutions, but usually only one or a few of these solu- 
tions satisfy the condition of maximima economy, or 
minimum expense. One of such cases is the selection 
of sizes of conductors in a network used for lighting 
or for power distribution. The total voltage drop 
between the generating station and the consmner is 
usually given, but this voltage drop may be arbitrarily 
divided, allowing part of it in the main feeders, another 
part in secondary feeders, and the remainder in the 
distributing conductors. If a large part of the total 
drop be allowed in the feeders, these feeders can be 
made of comparatively small cross section, but the 
distributing conductors would have to be compara- 
tively large, and vice versa. The problem becomes 
definite by adding the condition that the total weight 
of copper (or any other metal) be a minimum. This 
method of determining cross sections of conductors is 
appUed to a few practical cases in the problems below. 
However, the student must imderstand that the answer 
so obtained is often modified in practice for the follow- 
ing reasons: 

(a) A cross section which gives theoretically the least 
amoimt of metal may not be a standard size, and it is 
usually cheaper to use a standard conductor than to 
order a special size drawn and insulated. 


Digitized by LjOOQ IC 


(6) Even with standard sizes, it is very desirable to 
reduce the number in a given installation to a minimuTn, 
so as to avoid mistakes in repairs, and not to keep too 
large a stock. 

(c) The conductors must satisfy the condition that 
the current density shall not exceed a certain practical 
limit, owing to the danger of overheating, damage to 
insulation, and fire risk. 

(d) Larger conductors than are necessary for electrical 
requirements are often used for mechanical strength, or 
in anticipation of the growth of the system. 

Prob. I. — Prove that one conductor of length I is cheaper than 
two conductors of different cross section in series. Solution, — Let 
the cross sections of two conductors be qi and 52 respectively, and 
their lengths Zi, and fe = Z — Zi. Let the total permissible voltage 
drop be e, and the current i. K the unknown voltage drop in one 
conductor be «, that in the other must be e — x. Appljdng Eq. (3) 
of the preceding chapter, we get two conditions: x = iph/qi, and 
e — X = ip{l — li)/q%* The total volume of metal must be a 
minimum, or qik + qi(l — li) = min. Substituting the values of 
qi and 52 from the first two equations into the third, we get: 
u = ZiVx + (Z - Zi)V(e - x) = min. 

According to the general rule of the differential calculus, the partial 
derivatives of this expression with respect to h and x must be 
equated to zero. In this case both conditions lead to the same 
equation: x/h = (c — x)/(l — Zi), which means that the voltage 
drop per unit length must be the same in both conductors. But 
since the current is the same, this condition means that qi = 92. 
See the note at the end of the chapter. 

Prob. 2. — Solve Prob. 1 using the method of indeterminate 
coefficients. Solution. — The total voltage drop in the two con- 
ductors is e = ip (h/qi + U/qi)- The sum of the lengths being 
given we have Zi + Z2 = Z. Thus, multipljdng the first equation 
by an indeterminate coefficient Xi and the second equation by Xj, 
and adding to the given condition of minimum, we get: qj,i + qjl^ + 
Xi (Zi/gi + It/qi) + X2 (Zi + Z2) = min. In this equation, gi, Zi, ^2, 
Z2 can now be treated as independent variables. Equating to zero 

Digitized by VjOOQ IC 



the partial derivatives of the expression, the desired resnlt is 

Prob. 3. — Two conductors of lengths h and k, and cross sections 
qi and ^2, are to be connected in parallel across a source of voltage E. 
The total current i drawn by the two conductors is given and also 
the sum of the cross sections. Prove that the weight of metal is a 
minimum when k = k. Hint, — If the current through one con- 
ductor is Xf and through the other (i — x), we have, according to 
Eq. (7) in the preceding chapter: x = Eyqi/h^ and i — x = Eyqt/h* 

Prob. 4. — An electric light main (Fig. 8) has groups of lamps 
connected to it at different places. It is required to determine the 
cross sections ^1, ^2, ^'8, etc., in such a way that the total weight of 




' U 




■ ) 

> t 

1 i 


t'l it is in 

Fig. 8. — An electric street cable with house connections. 

metal in the conductors be a minimum. The applied voltage E 
is given, as well as all the currents. The total voltage drop in the 
conductor must not exceed 2 e, so that the voltage across the most 
remote lamps is E — 2e. Solution. — The voltage drop in one of 
the conductors, say the positive one, is 

e = pi:il/q (1) 

The condition for a minimum volume of metal is 

7=Sig = min (2) 

The problem is one in relative maxima and minima, so that we have: 
Zlq + \^Xil/q = min., (3) 

where X' is an indeterminate coefficient. Taking partial deriva- 
tives with respect to cross sections q, we obtain n equations of the 

h - xHVfc/g»2 = 0, 
or, solving for qk, 

g* = X • v^ (4) 

Substituting these values of q's in Eq. (1), we get: 

X = (p/e) XI VI (6) 

Knowing X, the cross sections are computed from Eq. (4). 

Digitized by VjOOQ IC 



[Chap. II. 

Prob. 5. — Show that in the preceding problem the current den- 
sity ik/qk decreases from the generator to the last consumer. 

Prob. 6. — Solve Prob. 4, substituting the condition of constant 
current density in place of Eq. (2). 

Prob. 7. — Let the currents consimied in the four groups of 
lamps in Fig. 8 be 40, 30, 50, and 27 amperes respectively. Let 
the distances be 1000, 3000, 450 and 1200 meters, and the applied 
voltage 228. The voltage across the fiuthest lamps must not be 
less than 220. Find the cross sections of the conductors and the 
total weight of copper, according to three different suppositions: 
(1) The weight is to be a minimum; (2) the current density is to 
be the same in all the sections; (3) the whole conductor is to be of 
a uniform cross section. For copper p = 0.018 ohm, when lengths 
are measured in meters and cross sections in square millimeters. 

Prob. 8. — Electric power is to be transmitted from a water- 
fall A (Fig. 9) to two towns C and D. Between A and B the line 

Fig. 9. — A branched transmission line. 

is common to both towns. Knowing the currents I'l and 12, the 
lengths of the lines, and the permissible voltage drop 2 e between 
A and C which is equal to that between A and Z), find the cross 
sections of the wires such that the total weight of the conductors 
will be a minimum. Solution, — The volume of the conductors is 

ql + qili + gjfe = min (6) 

Let the unknown drop of voltage between A and B be x. Then 
we have three conditions: 

e — X = pliii/qi = pkii/q^i (7) 

^^^ x = pZ(iV+t2)/(? (8) 

Substituting the values of 51, gj, and q from Eqs. (7) and (8) into 
Eq. (6) and differentiating with respect to x, we get after reduction: 

X l — x 

Digitized by VjOOQ IC 


From this equation x can be calculated, and subsequently the three 
unknown sizes of conductors found. 

Prob. 9. — The voltage at A (Fig. 9) is 1300 volts (for electric 
railway service); the voltage at C and at D must not be below 1150 
volts when the current consumption at C is 100 amperes, and at 
D 200 amperes. The distances are Z = 4 km.; h = 2 km.; k = 1.3 
km. Determine the most economical size of conductors and the total 
cost of copper at 40 cents a kilogram. 

Prob. 10. — Solve the preceding problem and compare the 
answers when, instead of minimum weight, the cross sections 
satisfy one of the following two conditions: (a) the current density 
is the same in the three branches; (b) the cross sections of the three 
branches are equal, and the voltage drop to neither of the points 
of consumption exceeds the prescribed limit. 

Prob. II. — Show that the result obtained in Prob. 8 may be 
interpreted as follows : Replace conductors BC and BD by a fictitious 
conductor BF (Fig. 9) of the same cross section as AB, and of a 

^ V^i + W \^l + ^2J 

The fictitious conductor is supposed to carry the same current 
(t'l + {2) as AB. The introduction of this conductor is convenient 
in numerical applications, because knowing L and I the voltage 
drop X is found from the proportion x :e = I : {1 + L), 

Prob. 12. — Show that in the case of more than two conductors 
radiating from point B (Fig. 9) the length of the fictitious conductor is 

^-s/W <"> 

the sunmiation being extended over all the branches except AB. 

Prob. 13. — Extend the solution of Prob. 8 to the case when the 
voltage drop to point C is different from that to D. Show that 
X must be determined from an equation of the fourth degree, and 
indicate how it may be solved by trial. 

Prob. 14. — Referring to Fig. 9, let there be n branches con- 
nected to AB at point B, and let a different voltage drop be pre- 
scribed for each branch. Show that the condition for a mi n imum 
weight of copper is 

1=2^ (12) 

Digitized by CjOOQ IC 


Prob. 15. — Show how to determine the sizes of the conductors 
in the branched transmission system shown in Fig. 10, to satisfy 
the condition of minimum copper or almninmn. The total drop 
of voltage to the five substations Z), E, G, H, and K must be the 
same. Solution, — Replace the three conductors 6, 7, 8 by an 

Fig. 10. — A branched transmission system. 

equivalent conductor 9, according to formula (11). Then add the 
length of 9 to that of 5, and replace the two conductors 4 and 
(5 + 9) by an equivalent conductor 10. Similarly, replace con- 
ductors (10 + 3>and 2 by the fictitious conductor 11, so that the 
given network of conductors is finally replaced by one simple con- 
ductor (1 + 11) canying the total current Si. The cross section 
of this conductor is found from the total given voltage drop, so that 
the voltage drop in conductor 1 can be calculated, and the cross 
sections of conductors 1 and 2 found. The remainder of the system 
is then treated in the same way, using the rest of the prescribed drop 
of voltage. 

NOTE. — In problems in which partial derivatives of a function 
are equated to zero in order to find a minimiun value, strictly speak- 
ing, it is necessary to investigate whether or not the values of the 
variables so found actually give a minimum. This can be done by 
investigating the sign of a small increment of the function, at 
and near the critical values of the variables. If the values of the 
variables actually correspond to a minimimi, then this increment is 
positive for any small variations of the variables whatsoever, 
positive or negative. In practice, this test is often omitted, be- 
cause usually one can satisfy oneself by a physical interpretation 
of the problem that no other stationary value is possible, except a 
true minimum. The same remark applies of course to a maximum. 

Digitized by VjOOQ IC 


However, should there be any doubt, such a test ought to be per- 
formed. For instance, in the case of Probs. 1 and 2 we have 

h/qi + h/Qi = a = Const.; (13) 

li + k = l; (14) 

u = hqi + kqi (15) 

Solving the first two equations for k and k, and substituting their 
values into the third equation, we get 

u = l(qi + q2) - aq\q2 (16) 

Let hqi and hq^ be some small increments of qi and q^. Then, 
according to Taylor's theorem for a function of two variables, we 

bu = (% Diw+5^2 Diu) + i (%2 1)iH+2 dqiSq2 DJ)tU+bq2^ D^hi) +etc. 
Taking partial derivatives of m, we get 

Diu = I — aq2\ D2U = I — aqi; Dihi = 0; A^ = 0; D1D2U = —a. 

du = {I — aq^ 5^1 + (^ — aq\) Sq2 — aSqiSq2. 

For the critical or stationary value of u the coeflSicients of dqi and 
8q2 must be equal to zero, and we get the result already obtained, 

qi = q2 = l/a, (17) 

so that 

du = —a8qi8q2 (18) 

Thus, near the stationary value, the sign of du depends upon whether 
5^1 and dq2 are of the same sign or of opposite signs. This can be 
determined from Eq. (13) by taking a complete variation: 

- QiSqi/qi' + kfiq2/q2^) -t (Sh/qi + Sfe/^z) = 0. . . (19) 

At the critical value of the function, qi = ^2, and, according to Eq. 
(14), 5Z1 + 5fe = 0. Therefore, in Eq. (19) the expressions in 
both parentheses are equal to zero separately. The expression in 
the first parenthesis gives 

«^i/«^2 = -kqiViW) (20) 

Thus, when Zi and k are positive, as is presupposed in this particu- 
lar physical problem, 8qi and 5q2 are of opposite signs. Therefore, 
in Eq. (18) 8u is always positive, and the values of qi and ^2 accord- 
ing to Eq. (17) actually correspond to the minimum value of the 

Digitized by LjOOQ IC 


In the preceding chapter the loss of voltage m the 
conductors is given, and therefore the waste of power 
in the form of i^ heat in the distributing network is 
fixed. Consequently, the most economical sizes of 
conductors are those which give a minimum weight of 
copper for the whole installation. Such conditions 
usually obtain in comparatively small local distribut- 
ing systems in which the voltage drop is not more than 
a few per cent of the applied voltage. 

In large transmission systems the permissible drop 
of voltage, or, to be more acciu'ate, the permissible loss 
of power in the line, is much greater and may vary 
within comparatively wide limits. In this case the 
proper selection of the size of conductors depends upon 
two factors: the cost of the line itself, and the cost of 
power lost in the line as Joulean heat Pr. The follow- 
ing table will make this clear: 







Size of 

Cost of wire, 

Ten per cent 
for interest, 
sinking fund 
and upkeep, 

Line loss, 


per year. 

Annual cost of 

power lost in 

the tine, 


Total operat- 
ing ooet of 
(3) +(5) 








Digitized by LjOOQ IC 


A certain amount of power is to be delivered at a 
considerable distance from the power house, by means 
of a high-tension transmission line. Different standard 
sizes of conductors which may be considered for this 
transmission are given in the first colunm in the table. 
The corresponding total cost of conductors is given 
in the second colunm. The third colunm gives the 
amount to be written off annually from the gross in- 
come, in order to pay for the interest, sinking fund, and 
repairs on the line. This item is estimated in this case 
to amount to about 10 per cent of the cost of the wire 
itself. The fourth column gives the values of the heat 
loss, tV, in the line; the larger the size of the wire the 
smaller this loss, being directly proportional to the 
resistance of the wire, in other words, inversely pro- 
portional to its cross section. In the fifth column will 
be found the actual cost of this line loss, in so far as 
the generation of power is concerned, or the loss of 
operating income. This item must also be subtracted 
from the gross income derived from the power 

Since the gross income may be considered fixed, the 
most economical size of conductor is that for which the 
total expense (column 6) is a minimum; in our case 
80 mm.2. By taking a larger size of wire, say 120 nam.^, 
the loss in the line is reduced from $2450 to $1633, 
but the interest and depreciation increase from $2500 
to $3750, so that the total cost of transmission is in- 
creased from $4950 to $5383. Reducing the size of 
wire to 50 mm.^ is not advisable either; the interest 
and depreciation in this case are comparatively small, 
only $1562, but the cost of power lost in the fine and 
the total cost of transmission are again higher. 

The table shows that the total cost is a minimum, 

Digitized by VjOOQ IC 



[Chap. III. 

when the items in columns 3 and 5 are approximately 
equal. It is proved below that this is the necessary- 
theoretical condition for a minimum operating cost. 
The table shows also that it is permissible in practice to 
deviate considerably from the most economical cross 
section, without appreciably increasing the cost of 
transmission. This fact is illustrated in Fig. 11: while 

Cro8S-8ectioii of Ck>nductoc 
Fig. 11. — Items in the operating cost of power transmission. 

the slope of both component curves is considerable, 
the resultant curve is quite flat near the point of mini- 
mum. Therefore, it is permissible in practice to mod- 
ify to a considerable extent the theoretically most 
economical cross section in order to satisfy some other 
important conditions. For instance, if the capital 
necessary to start the enterprise is limited, it may be 

Digitized by VjOOQ IC 


advisable to choose the 70 rnm.^ conductor instead 
of the 80 mm.2. On the contrary, if the capital is 
plentiful, and the demand for power may increase in 
the near future, it is better to take 90 mm.^ con- 

The most economical cross section is not always the 
one for which the total cost of transmission is a min- 
imum. It will be seen from the problems below that 
sometimes it is proper to design the line for a maximum 
of net income; again, in otTier cases, the line is to be so 
proportioned as to insiu'e the greatest percentage gain on 
the invested capital. The student should not expect 
that the solution of the problems given below would 
enable him to handle practical transmission projects 
in their entire complexity. These problems are in- 
tended only to elucidate an elementary economic aspect 
of the subject that can be reduced to mathematical 
maxima and minima. 

The total operating expense always includes large 
items independent of the size of conductors, for in- 
stance, the interest on the cost of transmission towers, 
the cost of patroUing the Une, etc. (see Prob. 2). There- 
fore, the total operating expense is a function of the 
cross section q of the conductors plus a constant quan- 
tity. When taking the derivative of the function in 
order to obtain a minimum, the constant term vanishes, 
so that the desired value of g is independent of the 
constant items of the operating cost. 

In order to simplify the mathematics and to avoid 
going into piu'ely electrical matters, all problems are 
solved for direct current, and not for alternating current 
transmission. With certain limitations, the results 
hold true for the latter, especially at power factor 
values near 100 per cent. 

Digitized by LjOOQ IC 


Prob. I. — In the table given above the cost C of wire may be 
represented by the formula 

C = aq, (1) 

where q is the cross section of the wire and a is a constant depend- 
ing upon the length of the line and the cost of wire per kilogram. 
The annual cost W of energy lost in the line can be expressed by 
the formula 

W^h/q, (2) 

where b is the annual expense when the cross section of the wire 
equals 1 mm.^. Prove that the most economical size of the wire is 
the one for which pC = TF, where p is the fraction of the original 
cost allowed for the interest, sinking fund, and upkeep of the wire 
per year. This relationship is known as Kelvin's law of economy. 
Solution. — The total cost of transmission per year is 

X = pC + W = paq + b/q (3) 

Equating to zero the first derivative with respect to g, we get 
pa — b/q^ = 0, or 

q = Vb/pa (4) 

Substituting this value in Eqs. (1) and (2), the desired relation is 

In Eq. (3) it is required to find the minimum value of the sum 
of two variables paq and b/q. The product of these variables is a 
constant quantity, pab, so that Theorem 1 in the Appendix applies, 
and we can write directly that 

paq = b/q, (5) 

or pC = TF. From Eq. (5) the value of q given in Eq. (4) follows 

Prob. 2. — In designing a transmission line it is found that the 
total annual cost X of transmission is the following function of the 
cross section q of the conductors: 

X^A + Bq^^ + C/q^ 

where A, B, C, w, and n are known constants. Determine the 
most economical value of q, 

Prob. 3. — A certain amount of power is to be transmitted by 
means of a high-tension direct-current line from a waterfall to a 
substation in a large city. Outside the city limits, the transmission 
is to be by means of bare wires on towers; within the city limits 

Digitized by VjOOQ IC 


it must be by means of underground cables. Determine the size 
of the bare wire and of the cable conductors under the condition that 
total annual expense for the interest, upkeep, and depreciation of the 
line be a minimum. Total voltage drop between the power house 
and the substation is fixed. Solviion. — Let the length and the 
cross section of the conductors for the overhead transmission be h 
and qi, those for the cable k and 52. Let the corresponding drops 
of voltage be ei and 62. Then, according to Eq. (3), Chapter I, 

qi = Ipli/ei'j qi = WsM, (6) 

and the condition must be fulfilled that 

61 + 62 = Const (7) 

Let the cost of the conductors per unit length and per unit cross 
section be ai and oz respectively. Then the annual cost of these 
conductors must satisfy the condition 

Piaikqi + ptaM2 = min., (8) 

where pi and p2 are the fractions of the original cost to be allowed 
as the annual expense. Substituting the values of qi and qt from 
Eq. (6) and omitting the common factor, />/, we get 

Piaik^/ei + P2fljt2^/e2 = min (9) 

The total differential of this expression must be equal to zero, or 
—piaiWei/e^ — p^nj^dez/e^ == 0. 

But from Eq. (7), cki + (fe = 0, so that, eliminating de\ and de^ 
from the preceding equation, we finally get 


Eq. (10) shows the ratio in which the total given voltage drop must 
be divided between the overhead and the underground conductors 
in order to reduce the expense to a minimum. Knowing 61 and 62, 
the corresponding values of qi and 52 are calculated from Eqs. (6). 
Prob. 4. — Power is to be transmitted to a city by bare overhead 
conductors for a distance of 10 km. and by underground cables 
within the city limits for a distance of 3 km. The total permissible 
voltage drop is 10 per cent of the generator voltage. What is the 
most economical per cent drop in the overhead line and in the cable 
if the unit price for the cable is 2.5 times that for the overhead 
conductor? Interest and amortization for the cable is estimated 

Digitized by VjOOQ IC 


at 6 per cent, while for the bare conductor 10 per cent should be 
allowed, due to its exposure. 

Prob. 5. — Extend the solution of Prob. 3 to the case of a line 
consisting of n sections in series, all having different given values of 
a, p and Z. Solution, — Eqs. (6) are of the same form, only there 
are n of them. Eq. (7) becomes 

Te* == 61 + ej + • • • + e» = Const. . . (7') 
Eq. (8) Ijecomes 


SfPifldkqk = min., (8') 

or, using Eqs. (6), 

2)p*o*Z»Vet = min (9') 


Eqs. (7') and (90 are solved by the general method of relative 
maxima and minima. Multiply Eq. (7') by an indeterminate 
coefficient X and add to Eq. (9'). Omitting the constant term, the 
result is: 


2) (PfcflAVcik + X«») = min. 

In this equation all e's are independent variables, so that taking 
partial derivatives we obtain n equations of the general form 
Pkadk^/ei? + X = 0. Consequently 

Ci : 62 : . . . : e»-i : e« = Zi Vplol : k Vp^ : k VpfOf : etc. 

This shows the ratio in which the total voltage drop must be divided 
among the different sections of the line, in order to have a minimum 
annual expense in interest, upkeep, and depreciation. 

Prob. 6. — Determine the most economical current density in 
a transmission line, taking into account the extra interest and 
amortization on the cost of the power plant. Solution, — The an- 
nual expenses to be considered in this case consist of three parts: 
<1) Interest and depreciation on the conductors. (2) Interest and 
depreciation on the excess equipment in the power house necessary 
to cover the losses in the conductors. (3) The cost of generation 
of extra energy to cover the loss in the line. 

Digitized by VjOOQ IC 



Let the current to be transmitted be i amperes, and let the volt- 
age drop in the Une be e. Then we have as before 

e = ilp/q, (11) 

so that the equipment in the power house must generate ei extra 
watts to cover the Hne loss. If the equipment is in operation T 
hours per year, the total annual expense X amounts to 

X = pioql + paftei + cTei = min (12) 

In this equation the term pioql represents the interest and deprecia- 
tion at the rate of 100 pi per cent on the cost aql of the line. The 
second term, pjm, represents the interest and depreciation at a 
rate of 100 p2 per cent on the extra rating of the machinery neces- 
sary to generate ei watts loss in the line, b being the cost of this 
machinery per watt of rating. The last term, cTei, represents the 
cost of the energy lost in the line, at the rate of c dollars per watt- 

Since the most economical current density S in the line con- 
ductors is required, the variables q and e in Eq. (12) must be ex- 
pressed through this density 

a-tVg (13) 

It follows from Eq. (11) that e = Zpd, and dividing both sides of 
Eq. (12) by a constant quantity pilf we finally obtain 

X/{pU) = (pia/p) (l/«) + (j)J> + cT)8 = min. . (13a) 

To find a mi n imum of cost, it is necessary to take the derivative of 
this expression with respect to «, and to equate it to zero. Noticing, 
however, that Theorem 1 in the Appendix applies in this case, we 
have (pia/p) (1/5) = (pjb + cT)S, from which 

«=-^^. ...... (14) 

VpJ) + cT 

This formula has been given an interesting interpretation by 
Hochenegg.* The quantity 

zi = v^Pia/p, (15) 

which enters into the numerator of the expression for 5, depends 
only upon the transmission line constants. Hochenegg calls Zi 
the " conductor factor " (Leitungszahl). The denominator, 

Z2 = VpJ, + cT, (16) 

* C. Hochenegg, Anordnung und Bemessung elektrischer Leitungen, 
Chap. Ill, p. 68. 

Digitized by LjOOQ iC 


depends only upon the operating conditions in the power house. 
Hochenegg gave it the name of "operating coeflSicient" (Betriebs- 
zahl). Thus, we get the simple relation that the most economical 
current density is equal to the ratio of the conductor factor to the 
operating coefficient: 

8 = zi/z2 (17) 

Prob. 7. — Determine the most economical current density in 
the line conductors for a power transmission system operating at 
full load for 3500 hoiu« a year, the cost of generation of energy being 
0.5 cent per kilowatt-hour ( = 1000 watt hoiu«) . Copper conductors 
are estimated to cost 360 dollars per metric ton (= 1000 kilograms), 
and the interest and depreciation on the conductors is assumed to 
be at a rate of 9 per cent per year. The excess capacity of the 
power house equipment costs 40 dollars per kilowatt of rating, 
and its interest and depreciation is at a rate of 10 per cent per 

Prob. 8. — At a waterfall a certain limited amount of energy 
is available in the form of a current i at E kilovolts. The cost of 
generation is ci cents per kilowatt-hour. The power is to be trans- 
mitted to a distance of } I km., to be sold there at a rate of Cs cents 
per kilowatt-hour. Determine the size of the line conductors such 
that the difference between the selling price of the energy delivered 
and the total cost of generation and transmission will be a maxi- 
mum. Sdviion, — Let the imknown cross section of the line be q] 
then the voltage drop in the line, in kilovolts, is 

e = iZp/g, (18) 

and the sum to be realized annually from the sale of the energy is 
02% (E — ilp/q) T, where T is the nimiber of hours per year during 
which the energy is used. The cost of generation per hour is aiE, 
and the interest and depreciation on the cost of the line conductors 
is paql. The condition to be satisfied is that 

at {E - ilp/q)T - CiiET - paql =^ max. . . (19) 

Omitting the constant terms, we find that the sum cjlRpTlq -|- pa^ 
must be a minimum. Since the product of these terms is a con- 
stant, the condition for the minimuni is cdHpT/q » po^, or 

'^. . . . (VM) 


= i\/^ 

Digitized by LjOOQ IC 


The corresponding current density is 

5 = 1 = ^:^ (196) 

Q Vc2T 

This expression is similar to Eq. (14). 

Prob. 9. — Solve the preceding problem when the cross section 
of the wire should be such as to yield the greatest possible per cent 
income on the capital aql invested in line conductors. Interpret 
the result in commercial terms. Hint. — Divide the left-hand side 
of expression (19) by aql and determine the value of q which makes 
it a maximum. 

Prob. 10. — Illustrate the difference between the conditions 
imposed in Probs. 8 and 9 on some numerical example, for instance, 
using the following data. The available power is 2000 kilowatts; 
it is to be transmitted at 22,000 volts to a town distant 20 km. 
The power is to be used 3000 hours a year, and to be sold at 5 cents 
per kilowatt-hour. The cost of one kilowatt-hour at the generating 
end of the line is 1.5 cents. For other data see Prob. 7. 

Prob. II. — As a modification of Prob. 8, let voltage E at the 
power house be unknown, but voltage E' at the receiving end be 
given. The total amount of power available at the waterfall is 
limited to P kilowatts. Find the most economical size of the 
conductors. Hint, — Eq. (19) becomes diWT - di (E' + e)T - 
paql = max. It contains three variables: i, e, and q. Two of 
these can be eliminated by means of Eq. (18) and the relation 
P = {E^ + e) I, or else the method of indeterminate coefficients 
can be used. 

Prob. 12, — Prove that the greatest amount of power which it is 
possible to deliver over a given line and at a given generator volt- 
age is obtained when one half of the generated power is lost in the 
line; in other words, when the efficiency of the line is 50 per cent. 
Illustrate this result by means of curves. Solution. — Let the 
values of the current and the voltage at the generator end be i and 
E; let the resistance of the line be r ohms. Then the delivered 
power is P ^ {E — ir) i = Ei — iV. This expression is a max- 
imum when E = 2 ir, or the voltage drop in the line la ir = iE. 
To current as abscissae plot curves of generated power, of power lost 
in the line, and of the delivered power. Take the current corre- 
sponding to the maximmn power as 100 amp., the generator voltage 
100 kilovolts; the resistance of the line must then be 500 ohms. 

Digitized by VjOOQ IC 


Prob. 13. — A power transmission company considers delivering 
electric power to a city, over an existing transmission line of 
resistance r. The receiver voltage E' is fixed. What amount of 
power would give a ma.ximTim net revenue to the company? Solvr 
tion, — Two cases have to be considered, according to whether or 
not the company can elsewhere dispose of the power not sold to the 

(1) There is a ready market for all the power available. Let 
% be the value of the current, Ci the cost of generating one kilowatt- 
hour, C2 the selling price of a kilowatt-hour. Then the condition for 
maximum net revenue is expressed as follows: 

CtEH - {ciE'i + ciih) = max (20) 

In this expression J^' is in kilovolts, z is in amperes, and r is in kilo- 
ohms. The first term represents the income, the second term the 
cost of the energy generated and transmitted, and the third term 
represents the loss in net revenue due to the loss of energy in the 
line; all three terms are in cents or in dollars per hour. Since by 
supposition there is a ready market for all the energy generated, the 
line loss is charged at the selling rate Cj. Taking a derivative with 
respect to i and equating it to zero, an expression is obtained from 
which I can be calculated. 

(2) The company cannot readily dispose of the excess power. 
In this case the loss in the Une cannot be charged at the rate C2, 
but only at the actual cost Ci. In other respects Eq. (20) remains 
the same. 

Prob. 14. — Apply the theory given in Prob. 13 to the following 
case: Energy is to be delivered at 33,000 volts over a line having a 
resistance of 27 ohms. Energy can be sold wholesale at 4 cents 
per kilowatt-hour, while the total cost of generation is 2.1 cents 
per kilowatt-hour. Find the most advantageous load in kilowatts 
to be transmitted under the two suppositions considered above. 
Plot curves showing the amount of gross and net revenue and the 
loss for various amounts of power transmitted. 

Prob. 15. — A number of villages are situated approximately 
along the circumference of a circle (Fig. 12) ; it is planned to supply 
them with electric power from a power house A situated at the 
center of the circle. Power is to be furnished by means of a cir- 
cular distributing line connected to radial feeders. The consumers 
and the current consumption may be considered to be uniformly 

Digitized by VjOOQ IC 



distributed over the ring conductor. The problem is to find the 
number n of feeders and the cross sections of the feeders and of 
the distributing Une such that the total cost of installation will be 
a minimum. The voltage drop both in the feeders and in the 


Fia. 12. — A centrally located power house with a number of 
consumers along the circle. 

distributing conductors is prescribed. Sdtdion, — Let the most 
economical number of feeders be n and let the prescribed voltage 
drop in each be ci. Let the total given current consumption be 
i, so that the ciurent per feeder is t'/n, and its cross section is 



The cost of each feeder per conductor is 

C/ = Z(ai(?i + 6i) + ci, (22) 

where ai, 6i, and Ci are certain constants. This expression is more 
complicated than those assumed in the preceding problems, because 
each new feeder means a new pole line, and extra switching and pro- 
tective apparatus at both ends. Therefore, the total cost of a 
feeder line consists of three items: laiQi, which is the cost of the 
conductor itself; Ibi, which is the cost of the pole line, independent 
of the cross section of the conductor; and Ci, which represents the 
cost of the switching and protective apparatus, etc., at both ends of 
the feeder. Each feeder, such as AB, supplies two sections, .BX and 

Digitized by VjOOQ IC 



BK\ of the distributing line. The uniformly distributed load 
between B and K may be considered concentrated in the middle 
between B and K. Therefore, the cross section of line KB is 

^^_ (i/2n)'{2^l/2n)p ^ ^23) 


and the cost of the distributing line over the whole circumference, 
per conductor, is 

Cd = 2irl*a^2 (24) 

The terms independent of the cross section, such as 61 and Ci, do not 
have to be considered here, because they vanish in the differentiation 
with respect to n. 

The total cost of installation is C« = nC/ + Cd, or 

Ct = n[l{aiqi + bi) + ci] + 2irla2q2. . . . (25) 

Substituting the values of qi and 32 from Eqs. (21) and (23) and 
equating to zero the first derivative of Ct with respect to n, the most 
economical niunber of feeders is determined. 

Prob. 16. — A high-tension transmission line AB (Fig. 13) runs 
side by side with a low-tension distributing line CD^ which supplies 


1 — 


— 1 




— 1 



— 1 



P ... "^ 

i — X — >i 
— ^ 1 - 


► 1^ 



A A' 

Fig. 13. — A distributing line connected to a transmission line 
through transformers. 

electric power to uniformly distributed consmners at an average 
rate of w watts per meter of its length. The low-tension line is 
connected to the high-tension line through transformers ti, <2, Ut 
etc. The power loss per unit length of the line must not exceed 
a given small fraction a of the delivered power. The cost of 
the high-tension line being fixed, it is required to find the best 
distance between the transformers, their size, and the cross section 
of the low-tension conductors, such as to make the total cost of the 
installation a miniTmiTn. SolvJtion, — Let the imknown distance 
between the transformers be 2 a; meters; then, neglecting a small loss, 
the size of each transformer \a2v)x watts, and its cost 2 aiwx + 61, 
where &i is a term which indicates that the price of a transformer is 

Digitized by VjOOQ IC 


not exactly proportional to its rating. The cost of the transform- 
ers per meter length of the line is aiw + (6i/2 x). Each transformer, 
such as <2, supplies power to the parts MN and MN' of the line. 
The power demand on the section MN of the line is xWy and the loss 
in the line must not be over axw. With a uniformly distributed 
load, the total power consimaption may be assumed to be concen- 
trated at one third of the length of the conductor; see Prob. 17 
below.* Hence, for two conductors, 

2iV = 2i^ •hxp/q = caw (26) 

Knowing the voltage E of the distributing line, current i can be 
calculated because xw = Ei, so that Eq. (26) gives 

»=i^ <^ 

The cost of the two conductors of the distributing line, per meter 
of its length, is 2 a^, so that the total cost of the installation, per 
meter of its length, is 

C^^arw + -+ 3^ =mm. . . . (28) 

Taking the first derivative with respect to x and equating it to 
zero, gives: 

2x'^^ ZoE^ ^' ^^^ 

whence the most economical distance between the transformers is 

2V 2a 

,^ (30) 

2 a^wp 

The student may well analyze this formula from an engineering 
point of view, making clear to himself the practical significance 
of the factors which enter into it, and of those which dropped 
out in the differentiation. It is also well worth while to multiply 
Eq. (29) by x and then compare it with Eq. (28). It will be seen 
that the total cost is a minimum when the distributing line costs 
twice as much as the excess cost 6i/2 x of the transformers. This 
result could be directly foreseen from Theorem 2 in the Appendix. 

* A uniformly distributed electric load is concentrated in the middle 
of the line for the purpose of determining the voltage drop, as in Prob. 
15, and is concentrated at one third the distance from the supply end 
for the purpose of determining the power loss, as in this problem. In 
the first case it is a matter of integrating a linear function ir of the 
current, in the second case a quadratic function ih of the current. 

Digitized by LjOOQ IC 


Prob. 17. — A distributing line furnishes a total current i which 
is uniformly consumed over its length Z. Prove that the power loss 
in the line, due to Joulean heat, is the same as if the total current 
consumption were concentrated at a distance J I from the generat- 
ing end. SolvMon, — At a distance x measured from the end of the 
line remotest from the generator, the current is ix/l; the power 
loss over an infinitesimal element dx of the line is (ix/I)^pdx/q. In- 
tegrating this expression between and I we get i^ • i Ip/q^ which 
proves the proposition. The curve of power loss per unit length is 
a parabola, so that the result could have been foreseen from a well- 
known theorem concerning the position of the center of gravity of an 
area bound by a parabola and the axis of abscissae. The proposi- 
tion proved in this problem is used in the solution of the preceding 

Digitized by LjOOQ IC 


Consider a long telegraph line with a battery of 
voltage El apphed at the sending end. The insula- 
tion of the line being imperfect, there is some leakage 
of current over the insulators and poles to the ground. 
Consequently, some current is drawn from the bat- 
tery even when the receiving end of the line is open. 
At each pole this leakage current depends upon the 
voltage between that particular point of the line and 
the ground. Since this voltage decreases from the 
sending to the receiving end of the line,. the leakage 
current per pole also decreases in the same way. The 
current which the wire carries at a particular point 
of the line is equal to the sum of the leakage currents 
between that point and the receiving end. Thus the 
line current decreases from a maximum near the bat- 
tery to zero at the opposite end of the line. 

When the receiver circuit is closed, the total current 
in the hne consists of two parts: the working current, 
which finally flows through the receiver relay, and the 
leakage current, which never reaches the other end. 
Under some conditions of bad weather and poor insu- 
lation, the leakage current of a long line may be much 
greater than the working current. 

Strictly speaking, the line current decreases in steps, 
from pole to pole. In order, however, to be able to 
apply the calculus, it is much easier to consider the 
leakage uniformly distributed over the line. With a 


Digitized by LjOOQ IC 


very long line, and a great many poles, the results so 
obtained are very close to the actual conditions. In 
the case of a leaky underground cable, the assumption 
may become perfectly correct. 

Let the resistance of the conductor be r ohms per 
kilometer, and let the conductance of the path be- 
tween the line and the ground be g mhos (ohm""^) per 
kilometer. With sufficient given conditions at the re- 
ceiver end, or at the sending end, or both, it is required 
to find the law according to which the current and the 
voltage vary along the line. 

For an infinitesimal length ds of the line, at a dis- 
tance s from the receiver end, we may write Ohm's 
law as follows: 

dE=irds, (1) 

where E is the voltage at that point between the line 
and the groimd. For the leakage current ovet the 
element ds we have 

di =Egds (2) 

The mathematical problem, therefore, is to solve 
Eqs, (1) and (2) so as to find E and i as functions of s. 
From Eq. (2), E = (l/g) (di/ds); taking a deriva- 
tive of both sides of this expression we get: dE/ds = 
(1/g) (dH/ds^). But from Eq, (1), dE/ds = ir, so that 

dH . ,oN 

d^ = ^ <3) 

This is the differential equation for the current i. 
In a similar manner, we obtain for the voltage E: 

For the purposes of solution of these equations, and 

Digitized by VjOOQ IC 


for numerical computations, it is convenient to intro- 
duce a factor 

m = Vffr, (5) 

so that the foregoing equations become 

B-"^- <«) 

f = -»*• w 

In other words, i and E are such functions of 8 that 
their second derivatives are proportional to the func- 
tions themselves. By a simple substitution the student 
will find that the exponential functions e"** and e"~* 
satisfy these equations, so that the most general solu- 
tion is 

i = Ae"^' +Be-^'] (8) 

E = Ce"*' + De-^'; (9) 

where A, By C and D are constants of integration to 
be determined from the given terminal conditions of 
the problem. For example, let it be desired to intro- 
duce the current I2 and the voltage E^ at the receiver 
end. Then for s = we have i = h and E = E2) 
and Eqs. (8) and (9) apphed to these points give two 
equations connecting the constants of integration. 
Two more equations are obtained by substituting the 
values of i and E from Eqs. (8) and (9) into either 
Eq. (1) or Eq. (2). 

While it is quite possible to solve the problem as 
outlined above, and to get the final result, both the 
steps and the final formulae are somewhat involved 
and unsynametrical; it is preferable to write the solu- 
tion of Eqs. (6) and (7) in the form of hyperbolic fimc- 

Digitized by VjOOQ iC 


tions. It will readily be seen that both cosh ms and 

sinhws satisfy these equations as partial solutions. 

Therefore the complete integrals of these equations 


i = h cosh ms -\'F sinh 7?zs; . . . (10) 

^ = ^2COshms +Xsinh7?zs; . . . (11) 

where F and K are constants of integration. The other 
two constants are denoted by U and J?2, because, by 
supposition, i = h when s = 0, and E = E2 when 
s = 0. Substituting the values of i and E from these 
equations into Eq. (1) we get 

7W^2sinh?ws+mX cosh ms=r72 cosh ?ws+ri^sinhw«. (12) 

Since this equation must hold true for all values of s, 
it must be an identity, so that 

F = mE2/r, (13) 

K = r/2/m. (14) 

It is convenient to introduce a fictitious resistance 

R=V^g, ...... (15) 

so that Eqs. (13) and (14) become 

F^E^/R (16) 

K = Rh (17) 

Eqs. (10) and (11) may now be written in th6 following 

i = h cosh ms + (E2/R) sinh ws, . . (18) 

E = E2 cosh ms + RI2 sinh ms. . . . (19) 

Several other expressions may be written for i and E, 
depending upon the particular problem in hseid, and 
upon whether the battery voltage is given. In all 
cases it is a question of properly selecting the form 

Digitized by LjOOQ IC 


of the hyperbolic functions and the constants of inte- 
gration. Tables of hyperboUc functions are available 
in a number of books, and computations are quite 
simple. ^ 

It is worthy of note that were the signs on one side 
of Eqs. (6) and (7) reversed, the solution would be in 
terms of ordinary circular sine and cosine. Such is 
the case in simple mechanical and electrical oscillations- 

Prob. I. — The resistance of a telegraph line is 7.2 ohms per kilo- 
meter, the insulation resistance to the ground is 0.2 megohm per 
km.; the length of the line is 450 km. The relay used at the re- 
ceiving end has a resistance of 280 ohms, and requires 0.11 amp. to 
operate it. Plot curves of E and i for the line from the receiving 
to the sending end. Hint. Ei = 280 X 0.11 = 30.8 volts. 

Prob. 2. — Show that expressions for F and K which are identical 
with Eqs. (16) and (17) are obtained by substituting the values of i 
and E from Eqs. (10) and (11) into Eq. (2), instead of into Eq. (1). 

Prob. 3. — Deduce expressions corresponding to Eqs. (18) and 
(19), using exponential in place of hjrperbolic functions. 

Prob. 4. — In order to determine the unknown conductance g of 
insulation, a telegraph line is opened at the receiver end, and simulta- 
neous readings ^f voltages are taken at both ends. Show how to 
determine g from these readings, knowing the resistance r of the 

Prob. 5. — In order to determine the unknown conductance g of 
insulation, a telegraph line is short-circuited at the receiver end 
and simultaneous readings of currents are taken at both ends. 
Show how to determine g from these reatdings, knowing the resist- 
ance r of the conductor. 

Prob. 6. — When distances s are measured from the sending 
end of the line, show that the sign on one side of Eqs. (1) and (2) 
must be reversed, while Eqs. (6) and (7) remain the same. 

Prob. 7. — Deduce expressions similar to Eqs. (18) and (19), in 
which 8 is measured from the sending end, and the constants of 
integration are expressed in terms of the current /i and the voltage 
El at the sending end. 

Prob. 8. — When the leakage conductance g is very small, m 
is also small, and R is very large. In Eqs. (18) and (19) replace 

Digitized by VjOOQ IC 


the hyperbolic functions by their expansions into infinite series, 
and show how to compute the voltage and the current at the send- 
ing end, neglecting all but two or three terms of the series. Con- 
sider the limiting case of ^ = 0, and show that the result could have 
been foreseen directly. 

Prob. 9. — Eq. (18) may be written in the form { = (/2/cosh $) 
cosh (ma + d), where the hjrperbolic angle $ is determined from the 
relationship tanh $ = E2 /(Rh). Write a corresponding transfor- 
mation for Eq. (19), using the same angle 6. 

Prob. 10. — A telegraph line 300 km. long has a resistance of 
6 ohm. per km. There are 15 poles per km. When the receiving 
circuit is open, and the sending end is connected to a 100-volt 
battery, the battery anmieter reads 0.13 amp. due to the leakage 
over the insulators. Compute the average insulation resistance 
per pole. 

Prob. II. — Extend the foregoing general theory to the case 
when the insulation conductance g varies along the line as a given 
continuous function of s, or when it has say two or three different 
constant values over certain portions of the line. Such a case arises 
when a different construction is adopted for different parts of the 
line, or when it rains over a portion of the line. 

Prob. 12. — As an application of Prob. 11, investigate the follow- 
ing practical case: A telegraph line extends between cities A and 
5, and identical batteries and relays are used at both ends. When 
it rains between A and a certain place C on the line, so that the 
leakage is increased on the portion AC, determine at which station, 
A or B, signals sent from the other end of the line are received 
more clearly. 

Note. — The theory given in this chapter may be extended to 
the case of long alternating-current transmission lines with dis- 
tributed capacity and inductance. Similar formulae apply here, 
except that hyperbolic functions of a complex variable are used. 
See the author's "Electric Circuit," page 208, and his article en- 
titled " Trigonometric Expressions for the Phenomena Occurring in 
Long-distance Transmission Lines," Electrical World, Vol. 66 (1915), 
p. 857; literature references are given in both places. 

Digitized by LjOOQ IC 



In a great many practical cases electric currents, 
voltages, and magnetic fluxes vary periodically with 
time, as shown in Fig. 14. 

The quantity under consideration, say the voltage 
across a line, increases, then decreases, is reversed, 


Fig. 14. — An alternating voltage. 

increases in the opposite direction, and decreases to 
zero again, as is shown in part by the curve ABCD. 
In some problems only the average value of the alter- 
nating quantity is required, and not the variable instan- 
taneous values. The average value A A' = ?/« is defined 
as such a value that the area between the curve ABC 
and the axis of abscissae is equal to the area of the 
rectangle AA'C'C. In other words, 

ACxya= r%dx (1) 


Digitized by LjOOQ IC 




Piob. I. — The electromotive force of an alternator varies ap- 
proximately according to a straight-line law (Fig. 15), the maximum 
value b being 3000 volts; what is the average value? 

Fig. 15. — An electromotive force which varies periodically accord- 
ing to a striaight-line law. 

Prob. 2. — A current varies with time as is shown in Fig. 16; 
it is equal to zero for p per cent of an alternation, and has a con- 
stant positive or negative value for the rest of the alternation. Let 


-100 -p- 



FiG. 16. — A rectangular alternating wave. 

the average value be equal to q per cent of the Tnfl.YiTT>iin> value. 
Derive an equation, and plot a curve between p and g. 














<- V 




Fig. 17. — A stepped curve of current or voltage. 

Piob. 3. — The electromotive force of an alternator varies 
according to a stepped curve shown in Fig. 17. Find the average 

Digitized by VjOOQ IC 



Prob. 4. — A current varies with time according to a sine curve 
(Fig. 18), the equation of which is { = / sin 2 irfty where / is the 
instantaneous maximum value, or the ampUtude of the wave, and 
/ is the frequency, or the number of complete cycles of the wave 

Fig. 18. — A sinusoidal current. 

per second. Find the average value of the current. Solution, — 
For a complete cycle the angle x = 2irft increases by 2t; hence, 
in terms of x, the base AC of a half-wave is equal to r. Therefore, 
according to Eq. (1), we have 

IT •!/ 

= ji dx = ^ J sin xdx =^2I, 

from which ia = (2/t) / = 0.6377. 

Prob. 5. — A non-sinusoidal curve, such as is shown in Figs. 
14 to 17, may be represented by an infinite series of sinusoidal 
expressions (Fourier's series) of certain frequencies or base lengths.* 
In a special case of a symmetrical curve this series becomes 

2/ = Ai sin « + As sin 3 a; + As sin 5 a; + etc., , . (2) 

where a; = 2 t/i^, as in the preceding problem. Show that the aver- 
age value of the ordinate 

y«=(2A)[Ai + iA, + JA5+ • • • ]. ... (3) 

Prob. 6. — What is the average value of a more general curve 

y = Ai sin (a; + ai) + As sin (3 x + as) + Aj sin (5 x + ai)+ etc., (4) 

where ai, aj, as, etc., are given angles? 

* For further details see the author's ** Experimental Electrical 
Engineering,'' Vol. II, p. 222, and his ''Electric Circuit," p. 42. 

Digitized by VjOOQ IC 




Prob. 7. — Find the average ordinate of the parabola (Fig: 19), 
given its width 2 a and its height 6. Solution, — Let the origin be 
at 0. According to a fundamental property of the parabola, we 
have for a point C, BC = p3B^, where p is a constant. The origin 

Fig. 19. — A parabolic curve. 

being at 0, the above equation can be written in the form (6 — j/) = 
yx^. The constant p is determined from the condition that x = a 
when 2/ = 0; hence, h = pa*, and the equation of the parabola is 

The average ordinate is determined from the expression 
a-2/a = 6jr(l-^^)(ia:=fa6, 

or ya = I &• This result could have been written directly, knowing 
that the area enclosed by a parabola is equal to two thirds of the 
area of the circumscribed rectangle. 

Prob. 8. — Solve the preceding problem for the case when the 
given curve is an hyperbola. 

Prob. 9. — Solve Prob. 7 when the given curve is a parabola 
of the nth degree. 

Prob. 10. — Solve Prob. 7 when the given curve is part of an 

Digitized by LjOOQ IC 


Prob. II. — Solve Prob. 7 when the given curve has a point of 
inflection as shown in curve No. 1 of Fig. 20. Consult some work 
on anal3rtic geometry for curves of such a shape. 

Prob. 12. — Arrange curves similar in general shape to those 
shown in No. 2 and No. 3 (Fig. 20), using portions of well-known 
curves, such as a parabola, an hyperbola, an ellipse, a sine-curve, 
etc., and determine their average values. 

Fio. 20. — Examples of wave forms. 

Beside the average value of an alternating current 
or voltage, another value is widely used in practice, 
known as the quadratic average or the mean effective 
value. In order to see its sipiificance, let a variable 
current i flow through a conductor of resistance r. 
The instantaneous energy dW converted into heat dur- 
ing an infinitesimal interval dt of time is dW = iV dt, 
and the heat energy developed during an interval of 
time T is 

iV dt. 

The value of a steady current, which, flowing through 
the same conductor for the same length of time T 
would develop the same amount of heat energy W as 

Digitized by LjOOQ IC 



[Chap. V. 

the variable current i, is called the effective value (i.) of 
the variable current i. Hence, by definition 

so that 


This is sometimes written in the form 

ien^^^fji'dt, .... (6) 

although the simpler form (5) is usually preferable. 

Prob. 13. — Calculate the effective value of the electromotive 
force shown in Fig. 15. Solution. — 

From similar triangles, y/b = x/a, so that 

ay,s =^r(bx/aydx = iai*, 

from which 2/, = 6/V3 = 1732 volts. 

Prob. 14. — Show that the result derived in the preceding prob- 
lem could be written directly, knowing that the area between a 
parabola and the tangent at the vertex is equal to one third of the 
area of the corresponding rectangle. 

















Fio. 21. — A trapezoidal alternating wave. 

Prob. 15. — Solve Prob. 2 for the effective value. 
Prob. 16. — Solve Prob. 3 for the effective value. 
Prob. 17. — Find the effective value of the trapezoidal wave 
shown in Fig. 21, and investigate the influence of the relative values 


Digitized by VjOOQ IC 


of percentages p, q^ and s. Also consider the cases when some or 
all of these percentages are equal to zero. 

Prob. 1 8. — Find the effective value of the sine wave shown in 
Fig. 18. Sdviion. — 

from which u = 1/^2. 

Prob. 19. — Prove that the effective value of the curve given in 
Prob. 5 is determined from the equation 

2y.^ = A,^ + At^ + A,^ + eUi (7) 

Hint, — The expression for y* consists of two kinds of terms, viz., 

the squares 

sin* J, sin* 3 x, sin* 5 x, etc., 

and the products 

sin x sin 3 x, sin a; sin 5 x, sin 3 a; sin 5 x, etc. 

The terms containing squares of sines are integrated as in the 
preceding problem. The terms containing products of sines vanish 
in integration, as is proved in the following problem. 

Prob. ao. — As an auxiliary theorem to the preceding problem 
prove that 

* f 'sin mx • sinnx • dx ^ 0, 

when m and n are odd integers. Solution, — The familiar relation, 
2 sin ?7ix sin na: = cos (w — n) a; — cos (m + n) x, gives 

r'sin mxBuinxdx = i f'cos (m — n)xdx — i j'cos (m + n)xdx 

__ 1 r sin (m — n) x "!^ _ lr siri(m + n)a: 'l»' _. a _ a _ a 
2L (m-n) Jo 2L {m + n) J)"" " "* 

Prob. 21. — Show that formula (7) holds true for the curve ex- 
pressed by Eq. (4). 

Prob. 22. — An irregular periodic curve is found to be approx- 
imately represented by the equation 

y = 20sinx + 6sin {Sx + 30°) + 4sin (5 x - 45**). 

Compute the mean quadratic ordinate by three methods: (a) Ana- 
l3rtically; (6) graphically in rectangular coordinates; (c) graphi- 
cally in polar coordinates. Show that the results are the same.* 

* For the graphical solutions see the author's "Electric Circuit," 
pp. 48 to 54. 

Digitized by LjOOQ IC 


Prob. 23. — Find the effective value of the parabolic curve 
given in Prob. 7. 

Prob. 24. — Find the effective value of the hyperbola the equa- 
tion of which is supposed to be known from Prob. 8. 

Prob. 25. — Find the effective values of the curves mentioned 
in Probs. 9 to 12. 

Prob. 26. — An electric circuit is connected to a source of par- 
tially rectified current, that is, a current consisting of a direct- 
current component to and of an alternating current J sin a;, so that 
the total current is y = io + /sinx. Find the average and the 
effective values of the current. 

Prob. 27. — Referring to the preceding problem, two ammeters 
are connected in the circuit: a hot-wire instrument which reads 
3ffective values, and a moving-coil permanent-magnet instrument 
which indicates average values. What relationship must exist 
between the values of to and / in order that the indications of both 
ammeters be the same? 

Digitized by LjOOQ IC 



The problems collected in this chapter are maxima 
and minima problems pertaining to the best dimensions 
of, and the best connections in, electric batteries, gen- 
erators, motors, and transformers. The mathematics 
in these problems is rather simple, and many of them 
can be solved even without differentiation, by apply- 
ing one of the theorems in the Appendix. These prob- 
lems are put at the end of the part on ''Electrical 
Engineering," because they presuppose more technical 
knowledge of electricity than the problems in the other 
chapters, and for this reason are perhaps of less general 

I J 


Fig. 22. — A series-parallel connection of electric cells. 

Prob. I. — It is required to arrange n cells of an electric battery 
in such a series-parallel combination (Fig. 22) as to obtain a max- 
imum current through a given external resistance R, The electro- 
motive force of each cell is e, its internal resistance is r. Solution, — 


Digitized by VjOOQ IC 


Let X cells be connected in series and n/x cells in parallel. The 
total electromotive force of the battery is xe; the total internal 

• TX TX^ 

resistance is — r- = — . Hence, the current 
n/x n 

i = -^ (1) 


Dividing the numerator and the denominator of this fraction by x 

i=p^ (2) 


X n 

This expression is a maximum when the denominator is a minimum. 
Theorem 1 in the Appendix applies to the denominator, because 

«.!:? = ^^ = Const., 
X n n 

so that it is a minimum when R/x = rx/n^ or 

fi = ^' (3) 


In other words, the current is a maximum when the external re- 
sistance is equal to the internal resistance of the battery. 

Prob. 2. — Derive result (3) by differentiating the dencmiinator 
of Eq. (2) directly, and show that it corresponds to a minimum and 
not to a maximimi. 

Prob. 3. — Deduce solution (3) by differentiating directly Eq. 
(1) ; investigate the sign of the second derivative. 

Prob. 4. — Prove that the efficiency of a battery, when it delivers 
a maximum current through a given external resistance, is only 
50 per cent. Note. — Efficiency is the ratio of the power Ei 
delivered to the external circuit, to the power Eqi developed in the 
battery. E is the terminal voltage of the battery, and Eq is its 
electromotive force on open circuit. 

Prob. 5. — A telegraph station has a battery of 364 Daniell cells, 
the internal resistance of each cell is 1 ohm, and the electromotive 
force of each cell, is 1.08 volt. What is the best way to connect 
these cells, so as to obtain a maximum current through a circuit 
of about 40 ohms resistance, and what will be the value of the cur- 
rent in the line and through each cell? 

Digitized by LjOOQ IC 


Prob. 6. — Assume the cells given in the preceding problem to 
be connected, first all in series, then in two parallel branches, in 
three branches, four branches, etc. Plot curves of line current 
and efficiency to number of branches as abscissae, and thus prove 
that in order to obtain a maximum current the cells must be con- 
nected in three parallel branches. 

Prob. 7. — Let it be required to solve Prob. 5 when 420 cells are 
available. According to Eq. (3), we find that 130 cells must be 
connected in series. Connecting three rows in parallel, the number 
of superfluous or idle cells is 420 — 3 X 130 = 30. It stands to 
reason, however, that a larger current could be obtained by using 
these 30 cells, in sets of 10, in series with each row, so that the true 
practical solution seems to be 140 X 3 = 420 cells. Explain this 
discrepancy and give a definite rule for using Eq. (3) in such cases. 

Prob. 8. — An adjustable electric heater is placed at the end of a 
long line of resistance r; the generator voltage is e. At what value 
of current does the heater receive a maximum of electric power? 
Solvtion, — Let the current be i; then the voltage drop in the line 
is ir and the voltage across the heater is e — ir. Thus, the power 
delivered to the heater is i (e — ir) and this expression must be a 
maximum. Multiplying this expression by a constant quantity r 
we see that 

ri+ (e — ri) = Const. 

hence Theorem 1 applies to it, and the power is a maximum when 
ri = e — rij or 

n = Je (4) 

In other words, the delivered power is a maximum when one half 
of the applied voltage is consumed in the line. 

Prob. 9. — Referring to the preceding problem, let the generator 
voltage be 100 volts and the resistance of the line 2.5 ohms. Plot 
values of power delivered to the heater and efficiency of the tranr- 
mission against current as abscissae. The resistance of the heater 
is adjustable between zero and 10 ohms. From these curves show 
that for maximum power, condition (4) must be satisfied. 

Prob. 10. — Define mathematically the curves obtained in the 
preceding problem, in terms of analytic geometry, and reduce their 
equations to the simplest possible form. 

Prob. II. — At what value of the counter-electromotive force is 
the power output of a series-wound direct-current motor a maxi- 

Digitized by VjOOQ IC 



mum, and what is its efficiency in this case, neglecting all the losses 
except the copper loss? Solution, — Let E be the voltage applied 
at the terminals of the motor and e its counter-e.m.f . The cur- 
rent through the armature is i = (E — e)/R, where B is the total 
resistance of the motor. The power output converted into mechan- 
ical work is et = e (^ — €)/R, because Ei is the input and i (E — e) 
is the copper loss in the armature itself. Since e + (^ — e) = 
Const., the condition for a maximum is e = -& — e, or e =« iE, The 
efficiency is 50 per cent. 1 

Prob. 12. — Solve Prob. 11, assuming that in addition to the 
copper loss there is a constant loss of K watts, to account for the 
iron loss and friction. 

Prob. 13. — A shunt-wound generator (Fig. 23) is operating at 
a constant terminal voltage e. The resistance of the armature is r, 
that of the shunt winding is R. At what current i is the efficiency 
of the machine a maximum? Neglect the friction and iron loss, 

-^ i 

Fig. 23. — A shunt-wound generator or motor. 

and assume the resistance of the field circuit to be practically con- 
stant within the operating limits. Sdution. — The field current is 
e/Rf so that the current through the armature is i + e/R. The 
loss of power in the field winding is e^/R, that in the armature is 
a + e/RYr, Hence, the efficiency of the machine is 


V = 

ei + eyR + (i + e/i2)V 

Instead of taking a derivative of this expression, we first divide its 
numerator and denominator by i, so as to make the numerator 
a constant. We thus obtain 

__ e 

" e + e^/ifii) + (i + e/RY (r/i) ' 

Digitized by LjOOQ IC 

Chap. Vl.l 



This expression becomes a maximum when the denominator is a 
minimiun. Omitting the constant terms, the denominator may be 
written in the form 

(^/Ri) (1 + r/R) + ri = min. 

The product of these two terms is constant, therefore they must 
be equal to each other. This condition gives an equation which 
can be solved for i. 

Prob. 14. — Referring to the preceding problem, let iiC be a con- 
stant iron and friction loss in watts. Show that the eifficiency is a 
maximum when the external current is 


+ 1. 


Prob. 15. — Plot a curve of per cent efficiency against line cur- 
rent for a twenty-kilowatt, shunt-wound generator operating at 
a terminal voltage of 550 volts. The armature resistance is 1.7 
ohms, the field resistance is 600 ohms, and the no-load losses 
K amount to 800 watts. Show that the curve is quite flat and 
that the maximum efficiency is at a current corresponding to expres- 
sion (5). 

Fig. 24. — A compound-wound generator. 

Prob. 16. — Modify the solution of Probs. 13 and 14 for a com- 
pound-wound generator (Fig. 24), the resistance of the series 
winding being ri. 

Prob. 17. — Modify the solution of Prob. 16 for a machine with 
a short shunt. This means that the shunt-field winding R (Fig. 24) 
is connected to the main circuit at B, instead of at A. 

Prob. 18. — A shunt-wound motor (Fig. 23) has an armature 
resistance r, a field resistance Rj and a no-load loss of K watts. 
What is the maximimi theoretical output P of the motor at a line 

Digitized by VjOOQ IC 


voltage of e volts (apart from the limits imposed by sparking and 
overheating)? SdtUion. — Let i be the armature current; the 
output is equal to the input minus the losses, so that 

Output P = (i + e/R) e - (^/R - iV - K), 

P^ie-Pr-K (6) 

This expression becomes a maximuni when i = i e/r (see Prob. 8). 

Prob. 19. — At what value of i does the motor in the foregoing 
problem operate at a maximum efficiency? Hint, — Use the total 
current I = i + e/R as the independent variable, and reduce the 
expression for efficiency to a sum of two terms the product of which 
is constant. 

Prob. 20. — Find the coordinates of the vertex of the parabola 
expressed by Eq. (6), and its parameter. Write the equation of 
this parabola in its simplest form taking the vertex for the origin. 

Prob. 21. — Prove that a transformer has a maximum efficiency 
at a load ei at which the copper loss iV is equal to the iron loss K. 

Solution. — Efficiency = — - — -— ; —, r-- i — 

output + copper loss + iron loss 

ei e 

ei + i^ + K e + ir + K/i 

The denominator must be a minimum, and since ir {K/i) = Const., 
ir = K/iy or iV == K, 

Prob. 22. — Is the preceding result modified if the losses are 
represented by the formula ih- + ai + K, where a is a constant? 
Note, — The power loss caused by the voltage drop across the 
brushes of a generator or a motor may be represented by the term ai. 

Prob. 23. — The efficiency of a transformer must be a maximum 
at three quarters of the rated load and must be equal to 95 per cent. 
What will its efficiency be at full load? 

Prob. 24. — The dimensions of the iron core of a transformer 
are given (Fig. 25), and also the available space for the copper, 
that is, the inside diameter Di of the winding and its outside diam- 
eter D2, Determine the thickness x of the inside coil (say the 
primary winding) and the thickness y of the outside secondary coil 
such as would make the total copper loss a minimum. Solution, — 
Let ni and Wa be the numbers of turns of the primary and of the 
secondary windings, ii and ii the corresponding currents, and qi and 
^2 the cross sections of the wire. The resistance of the primary 

Digitized by VjOOQ IC 

Chap. VI.] 



winding is pnnr (Di + x)/gi, where p is the resistivity of the material 
and TT (Di + x) is the mean length of one turn. Therefore the 
condition for minimum copper loss is 

punr (Di + x) I'lVgi H- pThJT (I>2 - y) i2V?2 = min. . . (7) 

The cross section of the primary wire is 51 = axh/ni, where h is the 
height of the coil and a is the " space factor " of the coil. It means 

K]/>t<-3^— > 


■— DiT 


Fig. 25. — A core-type transformer with concentric coils. 

that while the gross cross section of the coil is xh, only a part a is 
utiKzed for the copper, the rest being taken by the insulation and 
by the air spaces. The useful cross section axh comprises rii wires, 
hence the cross section of each wire is oah/ni. Similarly, q^ = 
ayh/n^. Substituting the values of qi and 52 in Eq. (7) and omit- 
ting all of the constant factors, we get 

riiHi^ (A + x)/x + nW (D2 — y)/y = min. 

It is proved in the theory of the transformer that Uiii = 112^2, 
with an accuracy sufficient for most practical purposes. Hence, 
the foregoing equation is simplified to 

A/X + 2)2/2/ = min (8) 

This expression, together with the condition 

a? + y = i (A - 2>i) = Const., .... (9) 

is sufficient for the solution of the problem. Differentiating these 
equations we get 

Dix-^ dx + D22/~* dy = 0; 
dx + dy ^ 0. 

Digitized by LjOOQ IC 



Substituting the value dy = — dx in the first of these equations, 
and rearranging the terms, we obtain 

x : 2/ = VWi : VdI (10) 

It is left to the student to prove that the values of x and y so ob- 
tained correspond to a minimuni of the function, and not to any 
other stationary value; see Note at the end of Chapter II. 

Prob. 25. — Using the results of the preceding problem, show 
that the diameter Do of the cyUndrical surface of separation (Fig. 25) 
between the primary and the secondary coils is the geometrical 
mean of the diameters Di and D2. 

Prob. 26. — Show that when the condition of minimum copper 
loss in Prob. 24 is satisfied, equal amounts of Joulean heat, iV, are 
developed in both windings. 

Prob. 27. — In a transformer Di = 36 cm., D2 = 68 cm., h = 
45 cm. The high-tension coil is on the outside; it has 2000 turns 
and carries a current of 10 amps. The low-tension current is 200 
amps. Calculate the cross sections of the primary and the second- 
ary wire, assuming the space factor to be 50 per cent for both coils. 

Prob. 28. — Solve Prob. 24 for the case when the space factors, 
ai and a2, of the two windings are different. 

Fig. 26. — A core-t3rpe transformer with flat coils. 

Prob. 29. — In practice, a cylindrical layer of insulation is inter- 
posed between the two coils. Let the thickness of this insulation be 
a cm. How will this new condition modify the solution of Prob. 24? 

Prob. 30. — Solve Prob. 24 when the two coils are placed side 
by side (Fig. 26) instead of being concentric. The unknown quan- 

Digitized by VjOOQ IC 

Chap. VI.l 



titles are the heights, x and y, of the coils. A layer of insulation, 
a, is to be taken into account between the coils. 

Prob. 31. — The magnetic circuit of an alternating-current ap- 
paratus (Fig. 27) has a cross section Qi on the part h of its length, 
and a cross section Q2 on the rest, k, of the length. Select Qi and Q2 

Fig. 27. — An iron core with two different cross sections. 

so that the hysteresis loss will be a minimum for a given total weight 
of the core. Solution. — The hysteresis loss per cubic cm. of the 
core may be assumed to be proportional to some nth power of 
the flux density B; therefore 

QiZi5i« + Q2feB2'' = min.; (11) 

the limiting condition being that the volume of iron 

V = liQi + l2Q2 = Const (12) 

Since the same flux * is produced in all parts of the core, we have 
$ = BiQi = B^2 = Const (13) 

Substituting the values of Bi and B2 from Eq. (13) into Eq. (11), 
and omitting the constant factors, we get 

0^+^^ = -- ^''^ 

Using the method of indeterminate coefficients, we have, from 
Eqs. (14) and (12), 

Digitized by LjOOQ IC 


or, taking partial derivatives, "^—Cn — 1) Zi/Qi" + xZi = 0, and 
— (n — 1) fe/Qa" + xfe = 0. From these expressions we readily 
find that Qi = Qj. 

Prob. 32. — Solve the preceding problem by direct substitution, 
eliminating one of the Q's, so as to avoid the use of partial deriva- 

Prob. 33. — Solve Prob. 31 with the requirement that the total 
iron loss consisting of hysteresis and of eddy currents be a minimum. 
Assimie that the loss per cubic cm. of the core can be represented by 
the formula oB^ + fiB\ Hint, Qih (oBr + ^BtT) + Qj^ {cSi"^ 
+ /8B2**) = min. 

Prob. 34. — A transformer core is given, and the space available 
for the windings is known. It is required to determine the flux 
density B in the core so as to have a minimum amount of loss 
for a given output. Solviion, — From the theory of the trans- 
former it is known that the copper loss at a given current and with a 
given volume of winding is inversely proportional to the square of 
the flux density; or 

Copper loss = CB-^, (15) 

where C is a constant. The iron loss may be assumed to be pro- 
portional to a certain power m of the magnetic induction, or. 

Iron loss = oB* (16) 


CB-^ + aB« = min., (17) 

which is easily solved for B, The following is an interesting inter- 
pretation of the result: Differentiating Eq. (17) and multiplying 
the result by B we obtain —2 CB"* + moB^ = 0. In other words, 
B must be selected so that 

Copper loss = i w [iron loss] (18) 

Prob. 35. — Referring to the foregoing problem let the iron loss 
be represented by the usual formula, oB*** + /8J5*, where the first 
term represents the hysteresis loss and the second term the eddy- 
current loss. Show that the total loss for a given output is a min- 
imimi, when the condition is fulfilled that 

Copper loss » 80 per cent of hysteresis loss + full eddy current loss. 
(Arnold's rule.) 

Digitized by LjOOQ IC 





« a >! 

Fio. 28. — A laminated magnetic 


Prob. 36. — A part of the magnetic circuit of an electrical appa- 
ratus (Fig. 28) is to be made of steel laminations, and must be 
of a given thickness h, and width a. The thickness of the insula- 
tion between the punchings is d. Determine the thickness x of the 
laminations such that the iron 
loss for a given alternating 
flux be a minimum. Solution. 
— By taking a few thick 
laminations the space lost for 
insulation is reduced, hence 
the flux density and the hys- 
teresis loss are reduced. On 
the other hand the eddy-cur- 
rent loss is increased, because 
this loss per imit volume is proportional to the square of the thick- 
ness of laminations. Thus, there must be a thickness x for which 
the total loss is a minimum. Let Bq equal the flux density and Vo 
the volume of iron if a piece of the same dimensions h and a were 
made out of solid iron. Let y be the space factor of the laminated 
core, that is, the part actually occupied by iron. For instance, if 
the insulation takes 10 per cent o{h,y=^ 0.90. In terms of x and 
5 we have 

y^x/{x + s). . (19) 

The actual flux density, B, in the laminated core is greater than Bo, 
namely, B = Bo/y. The actual volume of iron in the laminated core 
is F = Foi^, and is less than Vo. According to Prob. 33, the con- 
dition for a minimum iron loss is 

F(aB« + /3J5*)«min. 
Substituting the values of F and B in terms of y, we get ' 

Pyi-~ + Q2/I— = min., (20) 

where P and Q are known constants. 

Prob. 37. — In the preceding problem, let y «= 0.85, m = 1.6, 
n = 2, 8 = 0.05 mm., and Bo = 15,000 lines per sq. cm. What 
is the ratio of a to /S when the core satisfies the condition of min- 
imum iron loss? 

Digitized by LjOOQ IC 


In many practical problems the product of two 

variables must be constant, while their sum must be a 

minimum. The mathematical problem is therefore to 

find the values of u and v that satisfy the conditions 

uv = C = const. ; 

u + V = min. 

Substituting the value of v from the first equation into 
the second, we obtain 

u + C/u = min. 
Equating the first derivative to zero, we get 
1 _ C/u^ = 0, 

or t^ = VC; consequently v = VC. We thus arrive 
at the following result : 

Theorem 1. — When the jyroducl of two variables is 
constant, their sum is a minimum when the variables are 
equal to each other. 

Knowing this theorem, it is not necessary to differ- 
entiate in each particular problem, but the solution 
may be written directly. See, for example, Prob. 1 in 
Chapter III. 

Sometimes the variables satisfy a more general con- 

ym^n = c = const., (1) 

and the expression to be made a minimum is of the 

t/p + t;« = min., (2) 

where m, n, p and q are known constants. 


Digitized by LjOOQ IC 


The total differential of the first expression is equal to 
zero, because the expression itself is a constant; the 
total differential of the second expression is also equal 
to zero, because the expression must be a Tnin i mum. 
We thus have 

rnvT'^if du + nu'^tf'^ dv = 0; 
pu^^ du + qv'^^ dv = 0. 

Eliminating du and dv, and simplifying, we get 

^ = ^ (3) 

v'' np 

The values of u and v are obtained by solving Eqs. (1) 
and (3) as simultaneous equations. This result we 
shall call 

Theorem 2. — When two variables satisfy condition 
(1), and expression (2) must he a minimum, the values of 
the variables must satisfy Eq. (3). For example, Eq. 
(28) in Chap. Ill can be solved using this theorem. 

In some problems the sum of two variables is con- 
stant, and their product must be a maximum; for 
instance, when it is required to find a rectangle of max- 
imiun area, the perimeter being given. The conditions 

u +v = const. ; 
uv = max. 

Proceeding as in the proof of Theorem 1, we get 

Theorem 3. — When the sum of two variables is con- 
stant, their product is a maximum when the variables are 
equal to each other. 
In a more general case we may have 

u"^ + V'' = const., (4) 

uH"^ = max (5) 

Proceeding as in the case of Theorem 2, we derive the 
following theorem: 

Digitized by LjOOQ IC 


Theorem 4. — When two variables satisfy condition 
(4), and expression (5) miLst be a maximum, the valves of 
the variables must be such that 

^ = ^ (6) 

V" mq 

The student will find throughout the book a number of 
other problems on maxima and minima in which the 
actual differentiation can be avoided by using one of 
these four theorems. 

Strictly speaking, in order to complete the proofs 
it is necessary to show that the critical values of the 
variables found above correspond to a minimum or a 
maximum respectively. This may be done by consider- 
ing the sign of the second derivative, by interpreting 
the results geometrically, or perhaps by some special 
reasoning. - 

Digitized by LjOOQ IC 


Arnold, E. — " Die Gleichstrommaschine," 2 vols., Springer. 

Arnold, E. — " Die Wechflelstromtechnik," 5 vols., Springer. 

Barnett, S. J. — " Elements of Electromagnetic Theory," Macmillan. 

Bedell, F. and Crehore, A. — " Alternating Currents," McGraw-Hill. 

Benischke, a. — *' Die Wissenschaf tlichen Grundlagen der Elektro- 
technik," Springer. 

Campbell, A. — " The Electron Theory," Cambridge Univ. Press. 

Christie, C. V. — "Electrical Engineering," McGraw-Hill. 

Cohen, L. — "Formul® and Tables for the Calculation of Alter- 
nating-current Problems," McGraw-Hill. 

Fleming, J. A. — "The Propagation of Electric Currents," Van 

Franklin, W. S. — "Electric Waves," Macmillan. 

Franklin, W. S. and MacNutt, B. ^ — " Advanced Theory of Elec- 
tricity and Magnetism," Macmillan. 

Herzog, J. and Feldmann, C. P. — " Die Berechnimg Elektrischer 
Leitungsnetze in Theorie und Praxis," Springer. 

Karapetoff, v. — "The Electric Circuit," McGraw-Hill. 

Karapetoff, V. — " The Magnetic Circuit," McGraw-Hill. 

Kennelly, a. E. — "The Application of Hyperbolic Functions to 
Electrical Engineering Problems," London University Press. 

Koch, E. H., Jr. — " Mathematics of Applied Electricity," Wil^ & 

LaCour and Bragstad. — " Alternating Currents," Longmans, Green. 

Lamb, C. G. — " Alternating Currents," Edward Arnold. 

Orlich, E. — " Kapazitat und Induktivitat," C. Vieweg. 

Pender, H. — " Principles of Electrical Engineering," McGraw-Hill. 

Roessler, G. — " Die Femleitung von Wechselstr6men," Springer. 

Steinmetz, C. p. — "Theoretical Elements of Electrical Engineer- 
ing," McGraw-Hill. 

Steinmetz, C. P. — " Theory and Calculation of Alternating-Current 
Phenomena," McGraw-Hill. 

Steinmetz, C. P. — " Theory and Calculation of Transient Electric 
Phenomena and Oscillations," McGraw-Hill. 

Thomson, J. J. — " Elements of the Mathematical Theory of Elec- 
tricity and Magnetism," Van Nostrand. 

Thomson, J. J. — "Notes of Recent Researches in Electricity and 
Magnetism," Clarendon Press. 

Dwight, H. B. — " Constant Voltage Transmission," Wiley & Sons. 


Digitized by LjOOQ IC 

Digitized by LjOOQ IC 

Digitized by LjOOQ IC 

Digitized by LjOOQ IC 

Digitized by LjOOQ IC 






^ ^ -.•.•.•.••••Vi^ ',*.V.-